NCERT Exemplar for Class 11 Maths Chapter 4 - Principle of Mathematical Induction (Book Solutions)

Prove statements in examples 1 to 5, by using the principle of mathematical induction for all \[n \in N\], that:

Example 1: $1 + 3 + 5 + ... + (2n - 1) = {n^2}$

Ans: Given that $P(n):1 + 3 + 5 + ... + (2n - 1) = {n^2}$.

Put $n = 1$

$P(1):1 = {1^2}$

$P(1):1$

So, the given statement is true for $n = 1$ and $P(1)$ is true.

Assume $P(n)$ is true for a natural number $n = k$.

$P(k):1 + 3 + 5 + ... + (2k - 1) = {k^2}$                                                                        

Now, we should prove that  $P(k + 1)$ is true.

$P(k + 1):1 + 3 + 5 + ... + (2k - 1) + (2k + 1) = {k^2} + (2k + 1)$

$P(k + 1):1 + 3 + 5 + ... + (2k - 1) + (2k + 1) = {(k + 1)^2}$

Hence, $P(k + 1)$ is true when $P(k)$ is true.

So, by the principle of mathematical induction,

$P(n):1 + 3 + 5 + ... + (2n - 1) = {n^2}$

Example 2: $\sum\limits_{t = 1}^{n - 1} {t(t + 1) = \dfrac{{n(n - 1)(n + 1)}}{3}}$, for all natural numbers \[n \geqslant 2\].

Ans: Given that $P(n):\sum\limits_{t = 1}^{n - 1} {t(t + 1) = \dfrac{{n(n - 1)(n + 1)}}{3}}$.

Put $n = 2$

\[P(2):\sum\limits_{t = 1}^{2 - 1} {t(t + 1) = \sum\limits_{t = 1}^1 {t(t + 1)} } \]

\[P(2):1.2\]

\[P(2):\dfrac{{1.2.3}}{3}\]

\[P(2):\dfrac{{2(2 - 1)(2 + 1)}}{3}\]

Hence, the given statement is true for $n = 2$ and $P(2)$ is true.

Assume $P(n)$ is true for a natural number $n = k$.

$P(k):\sum\limits_{t = 1}^{k - 1} {t(t + 1) = \dfrac{{k(k - 1)(k + 1)}}{3}}$                                                                        

Now, we should prove that  $P(k + 1)$ is true.

$P(k + 1):\sum\limits_{t = 1}^{k + 1 - 1} {t(t + 1) = } \sum\limits_{t = 1}^k {t(t + 1)}$

$P(k + 1):\sum\limits_{t = 1}^{k - 1} {t(t + 1) + k(k + 1)}$

$P(k + 1):\dfrac{{k(k - 1)(k + 1)}}{3} + k(k + 1)$

$P(k + 1):k(k + 1)\left[ {\dfrac{{k - 1 + 3}}{3}} \right]$

\[P(k + 1):\dfrac{{k(k + 1)(k + 2)}}{3}\]

\[P(k + 1):\dfrac{{(k + 1)((k + 1) - 1)((k + 1) + 1)}}{3}\]

So, $P(k + 1)$ is true when $P(k)$ is true.

So, by the principle of mathematical induction,

$P(n):\sum\limits_{t = 1}^{n - 1} {t(t + 1) = \dfrac{{n(n - 1)(n + 1)}}{3}}$ for all natural numbers$n \geqslant 2$.

Example 3: $\left( {1 - \dfrac{1}{{{2^2}}}} \right).\left( {1 - \dfrac{1}{{{3^2}}}} \right)...\left( {1 - \dfrac{1}{{{n^2}}}} \right) = \dfrac{{n + 1}}{{2n}}$, for all natural numbers \[n \geqslant 2\].

Ans: Given that $P(n):\left( {1 - \dfrac{1}{{{2^2}}}} \right).\left( {1 - \dfrac{1}{{{3^2}}}} \right)...\left( {1 - \dfrac{1}{{{n^2}}}} \right) = \dfrac{{n + 1}}{{2n}}$.

Put $n = 2$

$P(2):\left( {1 - \dfrac{1}{{{2^2}}}} \right) = \dfrac{{2 + 1}}{{2(2)}}$

$P(2):\dfrac{3}{4}$

Hence, the given statement is true for $n = 2$ and $P(2)$ is true.

Assume $P(n)$ is true for a natural number $n = k$.

$P(k):\left( {1 - \dfrac{1}{{{2^2}}}} \right).\left( {1 - \dfrac{1}{{{3^2}}}} \right)...\left( {1 - \dfrac{1}{{{k^2}}}} \right) = \dfrac{{k + 1}}{{2k}}$                                     $...(1)$                                   

Now, we should prove that  $P(k + 1)$ is true.

$P(k + 1):\left( {1 - \dfrac{1}{{{2^2}}}} \right).\left( {1 - \dfrac{1}{{{3^2}}}} \right)...\left( {1 - \dfrac{1}{{{k^2}}}} \right).\left( {1 - \dfrac{1}{{{{(k + 1)}^2}}}} \right)$

From \[\left( 1 \right)\]

$P(k + 1):\dfrac{{k + 1}}{{2k}}.\left( {1 - \dfrac{1}{{{{(k + 1)}^2}}}} \right)$

$P(k + 1):\dfrac{{{k^2} + 2k}}{{2k(k + 1)}}$

$P(k + 1):\dfrac{{(k + 1) + 1}}{{2(k + 1)}}$

Hence, $P(k + 1)$ is true when $P(k)$ is true.

So, by the principle of mathematical induction,

$P(n):\left( {1 - \dfrac{1}{{{2^2}}}} \right).\left( {1 - \dfrac{1}{{{3^2}}}} \right)...\left( {1 - \dfrac{1}{{{n^2}}}} \right) = \dfrac{{n + 1}}{{2n}}$ for all natural numbers $n \geqslant 2$.

Example 4: ${2^{2n}} - 1$ is divisible by 3.

Ans: Given that $P(n):{2^{2n}} - 1$

Put $n = 1$

$P(1):{2^2} - 1$

$P(1):3$

It is divisible by 3.

Hence, the given statement is true for $n = 1$ and $P(1)$ is true.

Assume $P(n)$ is true for a natural number $n = k$.

$P(k):{2^{2k}} - 1$ is divisible by $3$

So, ${2^{2k}} - 1 = 3q$, where $q \in N$                   $...(1)$                                                 

Now, we should prove that  $P(k + 1)$ is true.

$P(k + 1):{2^{2(k + 1)}} - 1$

$P(k + 1):{2^{2k + 2}} - 1$

$P(k + 1):{2^{2k}}{.2^2} - 1$

$P(k + 1):{2^{2k}}.4 - 1$

$P(k + 1):{3.2^{2k}} + ({2^{2k}} - 1)$

From \[\left( 1 \right)\]

$P(k + 1):{3.2^{2k}} + 3q$

$P(k + 1):3({2^{2k}} + q)$

It is divisible by 3.

Hence, $P(k + 1)$ is true when $P(k)$ is true.

So, by the principle of mathematical induction,

$P(n):{2^{2n}} - 1$ is divisible by 3.

Example 5: $2n + 1 < {2^n}$, for all natural numbers \[n \geqslant 3\].

Ans: Given that $P(n):2n + 1 < {2^n}$.

Put $n = 3$

$P(3):2(3) + 1 < {2^3}$

$P(3):7 < 8$

Hence, the given statement is true for $n = 3$ and $P(3)$ is true.

Assume $P(n)$ is true for a natural number $n = k$.

$P(k):2k + 1 < {2^k}$                                           $...(1)$                                          

Now, we should prove that  $P(k + 1)$ is true.

$P(k + 1):2(k + 1) + 1$

$P(k + 1):2k + 1 + 2$

From $(1)$

$P(k + 1):2k + 1 + 2 < {2^k} + 2$

$P(k + 1):2k + 1 + 2 < {2^k} + 2 < {2^k}.2$

$P(k + 1):2k + 3 < {2^{k + 1}}$

Hence, $P(k + 1)$ is true when $P(k)$ is true.

So, by the principle of mathematical induction,

$P(n):2n + 1 < {2^n}$ for all natural numbers $n \geqslant 3$.

Example 6: Define the sequence ${a_1},{a_2},{a_3}...$ as follows:

${a_1} = 2,{a_n} = 5{a_{n - 1}}$ , for all natural numbers \[n \geqslant 2\].

(1)Write the first four terms of the sequence.

Ans: Given that ${a_1} = 2$ and ${a_n} = 5{a_{n - 1}}$.

${a_2} = 5{a_{2 - 1}}$

${a_2} = 5{a_1}$

${a_2} = 10$

${a_3} = 5{a_{3 - 1}}$

${a_3} = 5{a_2}$

${a_3} = 50$

${a_4} = 5{a_{4 - 1}}$

${a_4} = 5{a_3}$

${a_4} = 250$

(2)Use the principle of mathematical induction to show that the terms of the sequence satisfy the formula ${a_n} = 2.{5^{n - 1}}$ for all natural numbers.

Ans: Given that ${a_1} = 2$, ${a_n} = 5{a_{n - 1}}$ and $P(n):{a_n} = {2.5^{n - 1}}$.

Put $n = 1$

$P(1):{a_1} = {2.5^{1 - 1}}$

$P(1):{a_1} = 2$

Hence, the given statement is true for $n = 1$ and $P(1)$ is true.

Assume $P(n)$ is true for a natural number $n = k$.

$P(k):{a_k} = {2.5^{k - 1}}$                                                                                     

Now, we should prove that  $P(k + 1)$ is true.

$P(k + 1):{a_{k + 1}} = 5.{a_k}$

$P(k + 1):{a_{k + 1}} = 5.({2.5^{k - 1}})$

$P(k + 1):{a_{k + 1}} = {2.5^k}$

$P(k + 1):{a_{k + 1}} = {2.5^{(k + 1) - 1}}$

Hence, $P(k + 1)$ is true when $P(k)$ is true.

So, by the principle of mathematical induction,

$P(n):{a_n} = {2.5^{n - 1}}$ for all natural numbers.

Example 7: The distributive law from algebra says that for all real numbers $c,{a_1}$ and ${a_2},$ we have $c({a_1} + {a_2}) = c{a_1} + c{a_2}$.

Use this law and mathematical induction to prove that, for all natural numbers, $n \geqslant 2,$if $c,{a_1},{a_2},...,{a_n}$ are any real numbers, then $c({a_1} + {a_2} + ... + {a_n}) = c{a_1} + c{a_2} + ... + c{a_n}$.

Ans: Given that $P(n):c({a_1} + {a_2} + ... + {a_n}) = c{a_1} + c{a_2} + ... + c{a_n}$.

Put $n = 2$

$P(2):c({a_1} + {a_2}) = c{a_1} + c{a_2}$

Hence, the given statement is true for $n = 2$ and $P(2)$ is true.

Assume $P(n)$ is true for a natural number $k,$ where $k > 2$.

$P(k):c({a_1} + {a_2} + ... + {a_k}) = c{a_1} + c{a_2} + ... + c{a_k}$                 $...(1)$                             

Now, we should prove that  $P(k + 1)$ is true.

$P(k + 1):c({a_1} + {a_2} + ... + {a_k} + {a_{k + 1}})$

$P(k + 1):c(({a_1} + {a_2} + ... + {a_k}) + {a_{k + 1}})$

$P(k + 1):c({a_1} + {a_2} + ... + {a_k}) + c{a_{k + 1}}$

From \[\left( 1 \right)\]

$P(k + 1):c{a_1} + c{a_2} + ... + c{a_k} + c{a_{k + 1}}$

Hence, $P(k + 1)$ is true when $P(k)$ is true.

So, by the principle of mathematical induction,

$c({a_1} + {a_2} + ... + {a_n}) = c{a_1} + c{a_2} + ... + c{a_n}$ for all natural numbers, $n \geqslant 2,$if $c,{a_1},{a_2},...,{a_n}$ are any real numbers.

Example 8: Prove by induction that for all natural numbers n

$sin\alpha  + sin(\alpha  + \beta ) + sin(\alpha  + 2\beta ) + ... + sin(\alpha  + (n - 1)\beta ) = \dfrac{{sin(\alpha  + \dfrac{{n - 1}}{2}\beta )sin\left( {\dfrac{{n\beta }}{2}} \right)}}{{sin\left( {\dfrac{\beta }{2}} \right)}}$

Ans: Given that $P(n):\sin \alpha  + \sin (\alpha  + \beta ) + \sin (\alpha  + 2\beta ) + ... + \sin (\alpha  + (n - 1)\beta ) = \dfrac{{\sin (\alpha  + \dfrac{{n - 1}}{2}\beta )\sin \left( {\dfrac{{n\beta }}{2}} \right)}}{{\sin \left( {\dfrac{\beta }{2}} \right)}}$

Put $n = 1$

$P(1):\sin \alpha  = \dfrac{{\sin (\alpha )\sin \left( {\dfrac{\beta }{2}} \right)}}{{\sin \left( {\dfrac{\beta }{2}} \right)}}$

Hence, the given statement is true for $n = 1$ and $P(1)$ is true.

Assume $P(n)$ is true for a natural number $k$ .

$P(k):\sin \alpha  + \sin (\alpha  + \beta ) + \sin (\alpha  + 2\beta ) + ... + \sin (\alpha  + (k - 1)\beta ) = \dfrac{{\sin (\alpha  + \dfrac{{k - 1}}{2}\beta )\sin \left( {\dfrac{{k\beta }}{2}} \right)}}{{\sin \left( {\dfrac{\beta }{2}} \right)}}{\text{    }}...(1)$               

Now, we should prove that  $P(k + 1)$ is true.

$P(k + 1):\sin \alpha  + \sin (\alpha  + \beta ) + \sin (\alpha  + 2\beta ) + ... + \sin (\alpha  + (k - 1)\beta ) + \sin (\alpha  + k\beta )$

From \[\left( 1 \right)\]

$P(k + 1):\dfrac{{\sin (\alpha  + \dfrac{{k - 1}}{2}\beta )\sin \left( {\dfrac{{k\beta }}{2}} \right)}}{{\sin \left( {\dfrac{\beta }{2}} \right)}} + \sin (\alpha  + k\beta )$

\[P(k + 1):\dfrac{{\sin (\alpha  + \dfrac{{k - 1}}{2}\beta )\sin \dfrac{{k\beta }}{2} + \sin (\alpha  + k\beta )\sin \dfrac{\beta }{2}}}{{\sin \dfrac{\beta }{2}}}\]

\[P(k + 1):\dfrac{{\cos \left( {\alpha  - \dfrac{\beta }{2}} \right) - \cos \left( {\alpha  + k\beta  - \dfrac{\beta }{2}} \right) + \cos \left( {\alpha  + k\beta  - \dfrac{\beta }{2}} \right) - \cos \left( {\alpha  + k\beta  + \dfrac{\beta }{2}} \right)}}{{2\sin \dfrac{\beta }{2}}}\]

\[P(k + 1):\dfrac{{\cos \left( {\alpha  - \dfrac{\beta }{2}} \right) - \cos \left( {\alpha  + k\beta  + \dfrac{\beta }{2}} \right)}}{{2\sin \dfrac{\beta }{2}}}\]

\[P(k + 1):\dfrac{{\sin \left( {\alpha  + \dfrac{{k\beta }}{2}} \right)\sin \left( {\dfrac{{k\beta  + \beta }}{2}} \right)}}{{\sin \dfrac{\beta }{2}}}\]

\[P(k + 1):\dfrac{{\sin \left( {\alpha  + \dfrac{{k\beta }}{2}} \right)\sin (k + 1)\left( {\dfrac{\beta }{2}} \right)}}{{\sin \dfrac{\beta }{2}}}\]

Hence, $P(k + 1)$ is true when $P(k)$ is true.

So, by the principle of mathematical induction,

$\sin \alpha  + \sin (\alpha  + \beta ) + \sin (\alpha  + 2\beta ) + ... + \sin (\alpha  + (n - 1)\beta ) = \dfrac{{\sin (\alpha  + \dfrac{{n - 1}}{2}\beta )\sin \left( {\dfrac{{n\beta }}{2}} \right)}}{{\sin \left( {\dfrac{\beta }{2}} \right)}}$  for all natural numbers $n$.

Example 9: Prove by the principle of mathematical induction that $1 \times 1! + 2 \times 2! + 3 \times 3! + ... + n \times n! = (n + 1)! - 1$ for all natural numbers n.

Ans: Given that $P(n):1 \times 1! + 2 \times 2! + 3 \times 3! + ... + n \times n! = (n + 1)! - 1$.

Put $n = 1$

$P(1):1 \times 1! = (1 + 1)! - 1$

$P(1):1$

Hence, the given statement is true for $n = 1$ and $P(1)$ is true.

Assume $P(n)$ is true for a natural number $k$. 

$P(k):1 \times 1! + 2 \times 2! + 3 \times 3! + ... + k \times k! = (k + 1)! - 1{\text{         }}...{\text{(1)}}$      

Now, we should prove that  $P(k + 1)$ is true.

$P(k + 1):1 \times 1! + 2 \times 2! + 3 \times 3! + ... + k \times k! + (k + 1) \times (k + 1)!$

From \[\left( 1 \right)\]

\[P(k + 1):(k + 1)! - 1 + (k + 1) \times (k + 1)!\]

\[P(k + 1):(k + 1 + 1)(k + 1)! - 1\]

\[P(k + 1):(k + 2)(k + 1)! - 1\]

\[P(k + 1):(k + 2)! - 1\]

Hence, $P(k + 1)$ is true when $P(k)$ is true.

So, by the principle of mathematical induction,

$P(n):1 \times 1! + 2 \times 2! + 3 \times 3! + ... + n \times n! = (n + 1)! - 1$ for all natural numbers $n$.

Example 10: Show by the principle of mathematical induction that the sum Sn of the n term of the series ${1^2} + 2 \times {2^2} + {3^2} + 2 \times {4^2} + {5^2} + 2 \times {6^2}...$ is given by

${S_n}$ = 

$\dfrac{{n{{(n + 1)}^2}}}{2},$ if n is even

$\dfrac{{{n^2}(n + 1)}}{2},$ if n is odd

Ans: Given that $P(n):{S_n}$ = 

$\dfrac{{n{{(n + 1)}^2}}}{2},{\text{ if }}n{\text{ is even}}$

  $\dfrac{{{n^2}(n + 1)}}{2},{\text{ if }}n{\text{ is odd}}$ 

We know that any term ${T_n}$ of the series is:

${T_n}$ = 

${n^2}{\text{ if }}n{\text{ is odd}}$

$2{n^2}{\text{ if }}n{\text{ is even}}$ 

Put $n = 1$

$P(1):{S_1} = {1^2}$

$P(1):{S_1} = \dfrac{{1.2}}{2}$

$P(1):{S_1} = \dfrac{{{1^2}.(1 + 1)}}{2}$

Hence, the given statement is true for $n = 1$ and $P(1)$ is true.

Assume $P(n)$ is true for a natural number $k$. 

Case 1: When $k$ is odd, then $k + 1$ is even.

$P(k + 1):{S_{k + 1}} = {1^2} + 2 \times {2^2} + ... + {k^2} + 2 \times {(k + 1)^2}$      

$P(k + 1):{S_{k + 1}} = \dfrac{{{k^2}(k + 1)}}{2} + 2 \times {(k + 1)^2}$

$P(k + 1):{S_{k + 1}} = \dfrac{{(k + 1)}}{2}[{k^2} + 4(k + 1)]$

As $k$is odd, ${1^2} + 2 \times {2^2} + ... + {k^2} = {k^2}\dfrac{{(k + 1)}}{2}$

$P(k + 1):{S_{k + 1}} = \dfrac{{k + 1}}{2}[{k^2} + 4k + 4]$

$P(k + 1):{S_{k + 1}} = \dfrac{{k + 1}}{2}{(k + 2)^2}$

\[P(k + 1):{S_{k + 1}} = (k + 1)\dfrac{{{{[(k + 1) + 1]}^2}}}{2}\]

Hence, $P(k + 1)$ is true when $P(k)$ is true in this case.

Case 2: When $k$ is even, then $k + 1$ is odd.

$P(k + 1):{1^2} + 2 \times {2^2} + ... + 2.{k^2} + {(k + 1)^2}$      

$P(k + 1):\dfrac{{k{{(k + 1)}^2}}}{2} + {(k + 1)^2}$

As $k$is even, ${1^2} + 2 \times {2^2} + ... + 2{k^2} = k\dfrac{{{{(k + 1)}^2}}}{2}$

$P(k + 1):\dfrac{{{{(k + 1)}^2}(k + 2)}}{2}$

$P(k + 1):\dfrac{{{{(k + 1)}^2}((k + 1) + 1)}}{2}$

Hence, $P(k + 1)$ is true when $P(k)$ is true in this case.

Hence, $P(k + 1)$ is true when $P(k)$ is true for any natural number $k.$

So, by the principle of mathematical induction,

$P(n)$ is true for all natural numbers.

Example 11: Let $P(n):{2^n} < (1 \times 2 \times 3 \times ... \times n)$. Then the smallest positive integer for which $P(n)$ is true is

(a) 1

(b) 2

(c) 3

(d) 4

Ans: Correct option is d. 

Given that $P(n):{2^n} < (1 \times 2 \times 3 \times ... \times n)$.

Put $n = 1$

$P(1):2 < 1$

It is false.

Put $n = 2$

$P(2):{2^2} < 1 \times 2$

$P(2):4 < 2$

It is false.

Put $n = 3$

$P(3):{2^3} < 1 \times 2 \times 3$

$P(3):8 < 6$

It is false.

Put $n = 4$

$P(4):{2^4} < 1 \times 2 \times 3 \times 4$

$P(4):16 < 24$

It is true.

Hence, the smallest positive integer for which $P(n)$ is true is $4.$

Example 12: A student was asked to prove a statement P(n) by induction. He proved that P(k+1) is true whenever P(k) is true for all $k > 5 \in N$ and also that P(5) is true. On the basis of this he could conclude that P(n) is true.

(a) for all $n \in N$

(b) for all n>5

(c) for all $n \geqslant 5$

(d) for all n<5

Ans: Correct option is c.

Given that $P(5)$ is true and $P(k + 1)$ is true whenever $P(k)$ is true.

So, $P(n)$ will be true for all $n \geqslant 5$.

Example 13: If  is divisible by $\lambda$ for all  is true, then the value of $\lambda$ is……..:

Ans: Given that $P(n):{2.4^{2n + 1}} + {3^{3n + 1}}$

Put $n = 1$

$P(1):{2.4^{2 + 1}} + {3^{3 + 1}}$

$P(1):2 \times 64 + 81$

$P(1):209$

Put $n = 2$

$P(2):{2.4^{2(2) + 1}} + {3^{3(2) + 1}}$

$P(2):2048 + 2187$

$P(2):4235$

The H.C.F. of the numbers $209$ and $4235$ is $11.$

Thus, ${2.4^{2n + 1}} + {3^{3n + 1}}$ is divisible by $11.$

Therefore, $\lambda$ is $11.$

Example 14: If  is divisible by 64 for all  is true, then the least negative integral value of $k$ is…….:

Ans: Given that $P(n):{49^n} + {16^n} + k$

Put $n = 1$

$P(1):{49^1} + {16^1} + k$

$P(1):65 + k$

As $P(n)$ is divisible by $64,$ $k$ should be $- 1.$

$\Rightarrow P(1):65 - 1 = 64$ which is divisible by $64.$

Therefore, $k$ is $- 1.$

Example 15: State whether the following proof (by mathematical induction) is true or false for the statement:

$P(n):{1^2} + {2^2} + ... + {n^2} = \dfrac{{n(n + 1)(2n + 1)}}{6}$

Ans: Given statement is false.

Given that $P(n):{1^2} + {2^2} + ... + {n^2} = \dfrac{{n(n + 1)(2n + 1)}}{6}$.

Put $n = 1$

$P(1):{1^2} = \dfrac{{1(1 + 1)(2 + 1)}}{6}$

$P(1):{1^2} = \dfrac{{(2)(3)}}{6}$

$P(1):1$

Hence, the given statement is true for $n = 1$ and $P(1)$ is true.

Assume $P(n)$ is true for a natural number $n = k$.

$P(k):{1^2} + {2^2} + ... + {k^2} = \dfrac{{k(k + 1)(2k + 1)}}{6}$                          $...(1)$                                                             

Now, we should prove that  $P(k + 1)$ is true.

$P(k + 1):{1^2} + {2^2} + ... + {k^2} + {(k + 1)^2}$

From $(1)$

$P(k + 1):\dfrac{{k(k + 1)(2k + 1)}}{6} + {(k + 1)^2}$ 

L.H.S. is not equal to the R.H.S $P(k + 1):\dfrac{{(k + 1)((k + 1) + 1)(2(k + 1) + 1)}}{6}$.

So,  $P(n)$ is false.

EXERCISE 4.1

Question 1: Give an example of a statement $P(n)$ which is true for $n \geqslant 4$ but $P(1),\;P(2)$ and $P(3)$ are not true. Justify your answer.

Ans: Let $P(n)$ be the statement $3n < n!$.

For $n = 1$

$3n = 3$

$n! = 1$

Here, $3 > 1$

So, $P(1)$ is false.

For $n = 2$

$3n = 6$

$n! = 2$

Here, $6 > 2$

So, $P(2)$ is false.

For $n = 3$

$3n = 9$

$n! = 6$

Here, $9 > 6$

So, $P(3)$ is false.

For $n = 4$

$3n = 12$

$n! = 24$

Here, $12 < 24$

So, $P(4)$ is true.

For $n = 5$

$3n = 15$

$n! = 120$

Here, $15 < 120$

So, $P(5)$ is true.

$P(n)$ can be verified that $3n < n!$ for $n = 6,7,8,....$

So, the statement $P(n):3n < n!$ is true for all $n \geqslant 4$ but $P(1),P(2)$ and $P(3)$ are false.

Question 2: Give an example of a statement $P(n)$ which is true for all n. Justify your answer.

Ans: Let $P(n)$ be the statement:

$P(n):{1^3} + {2^3} + {3^3} + ... + {n^3} = \dfrac{{{n^2}{{(n + 1)}^2}}}{4}$

For $n = 1$

${1^3} = 1$

$\dfrac{{{1^2}{{(1 + 1)}^2}}}{4} = 1$

So, $P(1)$ is true.

For $n = 2$

${1^3} + {2^3} = 9$

$\dfrac{{{2^2}{{(2 + 1)}^2}}}{4} = 9$

So, $P(2)$ is true.

For $n = 3$

${1^3} + {2^3} + {3^3} = 36$

$\dfrac{{{3^2}{{(3 + 1)}^2}}}{4} = 36$

So, $P(3)$ is true.

Therefore, $P(n):{1^3} + {2^3} + {3^3} + ... + {n^3} = \dfrac{{{n^2}{{(n + 1)}^2}}}{4}$ is true for all $n$.

Question 3: ${4^n} - 1$ is divisible by 3, for each natural number n.

Ans: We have $P(n) = {4^n} - 1$

Put $n = 1$

$P(1) = {4^1} - 1$

$P(1) = 3$ 

It is divisible by 3.

Now, take $n = 2$

$P(2) = {4^2} - 1$

$P(2) = 15$ 

It is divisible by $3$.

Now, take $n = 3$

$P(3) = {4^3} - 1$

$P(3) = 63$ 

It is divisible by $3$.

Let $P(k) = {4^k} - 1$ be divisible by 3,

$\Rightarrow {4^k} - 1 = 3x$                                   $....(1)$

And also, 

$P(k + 1) = {4^{k + 1}} - 1$

From $(1)$

$P(k + 1) = 4(3x + 1) - 1$

$P(k + 1) = 12x + 3$ is divisible by 3.

$\Rightarrow P(k + 1)$ will be true whenever $P(k)$ is true.

Therefore, $P(n) = {4^n} - 1$ is divisible by 3 is true for each natural number $n$.

Question 4: ${2^{3n}} - 1$ is divisible by 7 for all the natural numbers n.

Ans: Given that $P(n) = {2^{3n}} - 1$

Put $n = 1$

$P(1) = {2^{3(1)}} - 1$

$P(1) = 7$ 

It is divisible by $7$.

So, $P(1)$ is true.

Assume $P(m)$ is true and as it is divisible by $7$, 

${2^{3m}} - 1 = 7m$.

Now, we need to prove that the statement given will be true for $n = m + 1$.

$P(m + 1) = {2^{3(m + 1)}} - 1$

$P(m + 1) = {2^{3m + 3}} - 1$

$P(m + 1) = {2^{3m}}{2^3} - 1$

As ${2^{3m}} - 1 = 7m$

We get $P(m + 1) = (7m + 1){2^3} - 1$

$P(m + 1) = 56m + 8 - 1$

$P(m + 1) = 7(8m + 1)$

Hence, ${2^{3(m + 1)}} - 1$ is divisible by $7$.

$\Rightarrow P(m + 1)$ is true.

Thus, we can say that $P(n) = {2^{3n}} - 1$ is divisible by $7$ for all the natural numbers $n$.

Question 5: ${n^3} - 7n + 3$ is divisible by 3 for all the natural numbers n.

Ans: Given that $P(n) = {n^3} - 7n + 3$

Put $n = 1$

$P(1) = 1 - 7(1) + 3$

$P(1) =  - 3$ 

It is divisible by $3$.

So, given statement is valid for the value $n = 1$ 

Assume that $P(n)$ is true for a natural number $n = k$.

$P(k) = {k^3} - 7k + 3$ is divisible by $3$ or ${k^3} - 7k + 3 = 3m,m \in N$         $....(1)$

Now, $P(k + 1) = {(k + 1)^3} - 7(k + 1) + 3$

$P(k + 1) = {k^3} + 1 + 3k(k + 1) - 7k - 7 + 3$

From equation $(1)$,

$P(k + 1) = 3m + 1 + 3k(k + 1) - 7$

$P(k + 1) = 3m + 3k(k + 1) - 6$

$P(k + 1) = 3m + 3[k(k + 1) - 2]$

$P(k + 1) = 3[m + k(k + 1) - 2]$ 

It is divisible by $3$.

Hence, $P(k + 1)$ will be true whenever $P(k)$ is true.

Thus, by using principle of mathematical induction, we can say that:

${n^3} - 7n + 3$ is divisible by $3$ for all the natural numbers $n$.

Question 6: ${3^{2n}} - 1$ is divisible by 8 for all natural numbers n.

Ans: Given that $P(n) = {3^{2n}} - 1$

Put $n = 1$

$P(1):{3^2} - 1$

$P(1) = 8$ 

$P(1)$ 

It is divisible by $8$.

So, statement is true for $n = 1$ and $P(1)$ is true.

Let $P(n)$ is true for a natural number $n = k$.

$P(k) = {3^{2k}} - 1$ is divisible by $8$ or ${3^{2k}} - 1 = 8m,m \in N$                $....(1)$

Now, we should prove that  $P(k + 1)$ is true.

$P(k + 1) = {3^{2(k + 1)}} - 1$

$P(k + 1) = {3^{2k}}{3^2} - 1$

From equation $(1)$,

$P(k + 1) = (8m + 1){3^2} - 1$

$P(k + 1) = (8m + 1)9 - 1$

$P(k + 1) = 72m + 9 - 1$

$P(k + 1) = 72m + 8$ 

$P(k + 1) = 8(9m + 1)$ 

It is divisible by $8$.

$P(k + 1)$ is true whenever $P(k)$ is true.

So, by the principle of mathematical induction, we can say that:

${3^{2n}} - 1$ is divisible by $8$ for all the natural numbers $n$.

Question 7: For any natural number n ${7^n} - {2^n}$ is divisible by 5.

Ans: Given that $P(n) = {7^n} - {2^n}$

Put $n = 1$

$P(1) = {7^1} - {2^1}$

$P(1) = 5$ 

It is divisible by $5$.

The statement given is true for $n = 1$ and $P(1)$ is true.

Assume that $P(n)$ is true for a natural number $n = k$.

$P(k) = {7^k} - {2^k}$ is divisible by $5$ or ${7^k} - {2^k} = 5m,m \in N$$....(1)$

Now, we should prove that  $P(k + 1)$ is true.

$P(k + 1) = {7^{k + 1}} - {2^{k + 1}}$

$P(k + 1) = {7^k}7 - {2^k}2$

$P(k + 1) = (5 + 2){7^k} - {2^k}2$

$P(k + 1) = (5){7^k} + (2){7^k} - {2^k}2$

$P(k + 1) = (5){7^k} + 2({7^k} - {2^k})$

From $(1)$,

$P(k + 1) = (5){7^k} + 2(5m)$

$P(k + 1) = 5({7^k} + 2m)$ 

It is divisible by $5$.

Hence, $P(k + 1)$ is true whenever $P(k)$ is true.

Therefore, by the principle of mathematical induction,

${7^n} - {2^n}$ is divisible by $5$ for any natural number $n$ .

Question 8: For any natural number n, ${x^n} - {y^n}$ is divisible by $x - y$, where x and y are any integers with $x \ne y$.

Ans: Given that $P(n) = {x^n} - {y^n}$

Put $n = 1$

$P(1) = x - y$

It is divisible by $x - y$.

Hence, the statement is true for $n = 1$ and $P(1)$ is true.

Let $P(n)$ is true for the natural number $n = k$.

$P(k) = {x^k} - {y^k}$ is divisible by $x - y$ or ${x^k} - {y^k} = m(x - y),m \in N$                $....(1)$

Now, we should prove that  $P(k + 1)$ is true.

$P(k + 1) = {x^{k + 1}} - {y^{k + 1}}$

$P(k + 1) = {x^k}x - {y^k}y$

$P(k + 1) = {x^k}(y - y + x) - {y^k}y$

$P(k + 1) = {x^k}y - {x^k}y + {x^k}x - {y^k}y$

\[P(k + 1) = {x^k}(x - y) + y({x^k} - {y^k})\]

From $(1)$,

$P(k + 1) = {x^k}(x - y) + ym(x - y)$

$P(k + 1) = (x - y)({x^k} + ym)$ which is divisible by $x - y$.

So, $P(k + 1)$ is true whenever $P(k)$ is true.

So, by the principle of mathematical induction,

${x^n} - {y^n}$ is divisible by $x - y$, where $x$ and $y$ are integers such that $x \ne y$.

Question 9: ${n^3} - n$ is divisible by 6, for each natural number $n \geqslant 2$.

Ans: Given $P(n) = {n^3} - n$

Put $n = 2$

$P(2) = {2^3} - 2$

 $P(2) = 6$ 

It is divisible by $6$.

So, the statement is true for $n = 2$ and $P(2)$ is true.

Assume $P(n)$ is true for a natural number $n = k$.

$P(k) = {k^3} - k$ is divisible by $6$ or ${k^3} - k = 6m,m \in N$                $....(1)$

Now, we should prove that  $P(k + 1)$ is true.

$P(k + 1) = {(k + 1)^3} - (k + 1)$

$P(k + 1) = {k^3} + 1 + 3k(k + 1) - (k + 1)$

$P(k + 1) = {k^3} + 1 + 3{k^2} + 3k - k - 1$

$P(k + 1) = ({k^3} - k) + 3k(k + 1)$

From $(1)$,

$P(k + 1) = 6m + 3k(k + 1)$ is divisible by $6$ as $k(k + 1)$ is even.

Thus, $P(k + 1)$ is true whenever $P(k)$ is true.

Therefore, by the principle of mathematical induction, we can say that:

${n^3} - n$ is divisible by $6$ for each natural number $n \geqslant 2$.

Question 10: $n({n^2} + 5)$ is divisible by 6, for each natural number n.

Ans: We have $P(n) = n({n^2} + 5)$

Substitute $n = 1$

$P(1) = 1({1^2} + 5)$

$P(1) = 6$ which is divisible by 6.

Hence, the given statement is true for $n = 1$ and $P(1)$ is true.

Let $P(n)$ is true for some natural number $n = k$.

$P(k) = k({k^2} + 5)$ is divisible by $6$ or $k({k^2} + 5) = 6m,m \in N$                $....(1)$

Now, we have to prove that  $P(k + 1)$ is true.

$P(k + 1) = (k + 1)[{(k + 1)^2} + 5]$

$P(k + 1) = (k + 1)[({k^2} + 2k + 6]$

$P(k + 1) = {k^3} + 3{k^2} + 8k + 6$

$P(k + 1) = ({k^3} + 5k) + 3{k^2} + 3k + 6$

$P(k + 1) = k({k^2} + 5) + 3({k^2} + k + 2)$

From $(1)$,

$P(k + 1) = 6m + 3({k^2} + k + 2)$ 

Now, ${k^2} + k + 2$ is always even.

Thus, $3({k^2} + k + 2)$ is divisible by 6 and hence $6m + 3({k^2} + k + 2)$ is divisible by 6.

Hence, $P(k + 1)$ is true whenever $P(k)$ is true.

Therefore, by the principle of mathematical induction,

$n({n^2} + 5)$ is divisible by 6, for each natural number $n$.

Question 11: ${n^2} < {2^n}$ for all natural numbers $n \geqslant 5$.

Ans: Given that $P(n) = {n^2} < {2^n}$

Put $n = 5$

$P(5):{5^2} < {2^5}$

$P(5):25 < 32$

 It is true.

Hence, the given statement is true for $n = 5$ and $P(5)$ is true.

Let $P(n)$ is true for a natural number$n = k$.

$P(k) = {k^2} < {2^k}$                                                                        $....(1)$

Now, we should prove that  $P(k + 1)$ is true.

$P(k + 1):{(k + 1)^2} < {2^{(k + 1)}}$

From above,

$P(k + 1):{k^2} + 2k + 1 < {2^k} + 2k + 1$                                     $....(2)$

 Let ${2^k} + 2k + 1 < {2^{k + 1}}$                                                     $....(3)$

${2^k} + 2k + 1 < {2^k}(2)$

$2k + 1 < {2^k}$

It is true for all $k > 5$.

From $(2)$ and \[(3)\] , we get ${(k + 1)^2} < {2^{k + 1}}$.

Hence, $P(k + 1)$ is true whenever $P(k)$ is true.

Thus, by the principle of mathematical induction, we can say that:

${n^2} < {2^n}$ for all natural numbers $n \geqslant 5$.

Question 12: $2n < (n + 2)!$ for all the natural numbers n.

Ans: Given that $P(n) = 2n < (n + 2)!$

Put $n = 1$

$P(1):2 < 3!$

$P(1):2 < 6$

 It is true.

Hence, $P(1)$ is true.

Let $P(n)$ is true for a natural number $n = k$.

$P(k) = 2k < (k + 2)!$                                                                        $....(1)$

Now, we should prove that  $P(k + 1)$ is true.

$P(k + 1):2(k + 1) < (k + 1 + 2)!$

$P(k + 1):2(k + 1) < (k + 3)!$

From $(1)$,

$2(k + 1) = 2k + 2 < (k + 2)! + 2$                                                   $....(2)$

Now, let $(k + 2)! + 2 < (k + 3)!$                                                     $....(3)$

$2 < (k + 3)! - (k + 2)!$

$2 < (k + 2)![k + 2]$

It is true for any natural number.

From $(2)$ and \[(3)\] , we get $2(k + 1) < (k + 3)!$

Hence, $P(k + 1)$ is true whenever $P(k)$ is true.

So, by the principle of mathematical induction, we can say that:

$P(n)$ is true for any natural number $n$.

Question 13: $\sqrt n  < \dfrac{1}{{\sqrt 1 }} + \dfrac{1}{{\sqrt 2 }} + ... + \dfrac{1}{{\sqrt n }}$, for all natural numbers $n \geqslant 2$.

Ans: Given that $P(n) = \sqrt n  < \dfrac{1}{{\sqrt 1 }} + \dfrac{1}{{\sqrt 2 }} + ... + \dfrac{1}{{\sqrt n }}$

Put $n = 2$

$P(2):\sqrt 2  < \dfrac{1}{{\sqrt 1 }} + \dfrac{1}{{\sqrt 2 }}$ 

 $1.414 < 1.707$

It is true.

Put $n = 3$

$P(3):\sqrt 3  < \dfrac{1}{{\sqrt 1 }} + \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 3 }}$ 

$1.732 < 2.284$ 

It is true.

Let $P(k) = \sqrt k  < \dfrac{1}{{\sqrt 1 }} + \dfrac{1}{{\sqrt 2 }} + ... + \dfrac{1}{{\sqrt k }}$ is true.

Add $\sqrt {k + 1}  - \sqrt k $ on both sides:

\[\sqrt k  + \sqrt {k + 1}  - \sqrt k  < \dfrac{1}{{\sqrt 1 }} + \dfrac{1}{{\sqrt 2 }} + ... + \dfrac{1}{{\sqrt k }} + \sqrt {k + 1}  - \sqrt k \]

We know that $\sqrt {k + 1}  - \sqrt k  = \dfrac{{\left( {\sqrt {k + 1}  - \sqrt k } \right)\left( {\sqrt {k + 1}  + \sqrt k } \right)}}{{\left( {\sqrt {k + 1}  + \sqrt k } \right)}}$

$\Rightarrow \sqrt {k + 1}  < \dfrac{1}{{\sqrt 1 }} + \dfrac{1}{{\sqrt 2 }} + ... + \dfrac{1}{{\sqrt k }} + \dfrac{1}{{\sqrt {k + 1} }}$

Hence, $P(k + 1)$ is true whenever $P(k)$ is true.

So, by the principle of mathematical induction, we can say that:

$\sqrt n  < \dfrac{1}{{\sqrt 1 }} + \dfrac{1}{{\sqrt 2 }} + ... + \dfrac{1}{{\sqrt n }}$ for all natural numbers $n \geqslant 2$.

Question 14: $2 + 4 + 6 + ... + 2n = {n^2} + n$ for all natural numbers n.

Ans: Given that $P(n):2 + 4 + 6 + ... + 2n = {n^2} + n$

Substitute different values for $n$.

$P(0):0 = {0^2} + 0$

 It is true.

$P(1):2 = {1^2} + 1$ 

It is true.

$P(2):2 + 4 = {2^2} + 2$ 

It is true.

$P(3):2 + 4 + 6 = {3^2} + 3$

 It is true.

Let \[P(k):2 + 4 + 6 + ... + 2k = {k^2} + k\] is true.

So, \[P(k + 1):2 + 4 + 6 + ... + 2k + 2(k + 1) = {k^2} + k + 2k + 2\]

\[ \Rightarrow P(k + 1):({k^2} + 2k + 1) + (k + 1)\]

\[ \Rightarrow P(k + 1):{(k + 1)^2} + (k + 1)\]

Hence, $P(k + 1)$ is true when $P(k)$ is true.

So, by the principle of mathematical induction,

$2 + 4 + 6 + ... + 2n = {n^2} + n$ is true for all natural numbers $n$.

Question 15: $1 + 2 + {2^2} + ... + {2^n} = {2^{n + 1}} - 1$ for all natural numbers n.

Ans: Given that $P(n):1 + 2 + {2^2} + ... + {2^n} = {2^{n + 1}} - 1$

Put $n = 1$

$P(1):1 + {2^1} = {2^{1 + 1}} - 1$ 

$P(1):3$

It is true.

Assume that $P(n)$ is true for some natural number $n = k$.

\[P(k):1 + 2 + {2^2} + ... + {2^k} = {2^{k + 1}} - 1\]                                    $...(1)$

Now, we should prove that $P(k + 1)$ is true.

\[P(k + 1):1 + 2 + {2^2} + ... + {2^k} + {2^{k + 1}}\]

From $(1)$

\[ \Rightarrow P(k + 1):{2^{k + 1}} - 1 + {2^{k + 1}}\]

\[ \Rightarrow P(k + 1):(2){2^{k + 1}} - 1\]

\[ \Rightarrow P(k + 1):{2^{(k + 1) + 1}} - 1\]

Hence, $P(k + 1)$ is true when $P(k)$ is true.

So, by the principle of mathematical induction,

$1 + 2 + {2^2} + ... + {2^n} = {2^{n + 1}} - 1$ is true for all natural numbers $n$.

Question 16: $1 + 5 + 9 + ... + (4n - 3) = n(2n - 1)$ for all natural numbers n.

Ans: Given that $P(n):1 + 5 + 9 + ... + (4n - 3) = n(2n - 1)$

Put $n = 1$

$P(1):1 = 1(2 \times 1 - 1)$ 

$P(1):1$

It is true.

Assume that $P(n)$ is true for a natural number $n = k$.

\[P(k):1 + 5 + 9 + ... + (4k - 3) = k(2k - 1)\]                                    $...(1)$

Now, we should prove that $P(k + 1)$ is true.

\[P(k + 1):1 + 5 + 9 + ... + (4k - 3) + [4(k + 1) - 3]\]

From $(1)$

\[ \Rightarrow P(k + 1):2{k^2} - k + 4k + 4 - 3\]

\[ \Rightarrow P(k + 1):2{k^2} + 3k + 1\]

\[ \Rightarrow P(k + 1):(k + 1)(2k + 1)\]

\[ \Rightarrow P(k + 1):(k + 1)[2(k + 1) - 1]\]

Hence, $P(k + 1)$ is true when $P(k)$ is true.

So, by the principle of mathematical induction,

$1 + 5 + 9 + ... + (4n - 3) = n(2n - 1)$ is true for all natural numbers $n$.

Question 17: A sequence ${a_1},{a_2},{a_3}...$ is defined by letting ${a_1} = 3$ and ${a_k} = 7{a_{k - 1}}$ for all natural numbers $k \geqslant 2$. Show that ${a_n} = 3.{7^{n - 1}}$ for all natural numbers.

Ans: Given that ${a_k} = 7{a_{k - 1}}$

$\Rightarrow {a_2} = 7 \times {a_1}$

$\Rightarrow {a_2} = 7 \times 3$

$\Rightarrow {a_2} = {3.7^{2 - 1}}$

$\Rightarrow {a_3} = 7 \times {a_2}$

$\Rightarrow {a_3} = {7^2} \times {a_1}$

$\Rightarrow {a_3} = 49 \times 3$

$\Rightarrow {a_3} = {3.7^{3 - 1}}$

$\Rightarrow {a_4} = 7 \times {a_3}$

$\Rightarrow {a_4} = {7^2} \times {a_2}$

$\Rightarrow {a_4} = {7^3} \times {a_1}$

$\Rightarrow {a_4} = {3.7^{4 - 1}}$

$\Rightarrow {a_n} = 7{a_{n - 1}}$

$\Rightarrow {a_n} = {3.7^{n - 1}}$

Let ${a_m} = 7{a_{m - 1}}$

$\Rightarrow {a_m} = {3.7^{m - 1}}$ be true.

$\Rightarrow {a_{m + 1}}$ is true when ${a_m}$ is true.

So, by the principle of mathematical induction,

${a_n} = {3.7^{n - 1}}$ is true for all natural numbers $n$.

Question 18: A sequence ${b_0},{b_1},{b_2}...$ is defined by letting ${b_0} = 5$ and ${b_k} = 4 + {b_{k - 1}}$ for all natural numbers k. Show that ${b_n} = 5 + 4n$ for all natural numbers n using mathematical induction. 

Ans: Given that ${b_n} = 5 + 4n$

Put $n = 0$

${b_0} = 5$

Put $n = 1$

${b_1} = 5 + 4$ 

${b_1} = 9$

As ${b_0} = 5$

$\Rightarrow {b_1} = 4 + {b_0} = 4 + 5 = 9$

Thus, $P(1)$ is true.

Assume that $P(n)$ is true for a natural number $n = m$.

$P(m):{b_m} = 5 + 4m$                                    $...(1)$

Now, we should prove that $P(m + 1)$ is true.

\[P(m + 1):{b_{m + 1}} = 5 + 4(m + 1)\]

As ${b_k} = 4 + {b_{k - 1}}$

\[ \Rightarrow {b_{m + 1}} = 4 + {b_{m + 1 - 1}}\]

\[ \Rightarrow {b_{m + 1}} = 4 + {b_m}\]

From $(1)$

\[ \Rightarrow {b_{m + 1}} = 4 + 5 + 4m\]

\[ \Rightarrow {b_{m + 1}} = 5 + 4(m + 1)\]

Hence, $P(m + 1)$ is true when $P(m)$ is true.

So, by the principle of mathematical induction,

$P(n)$ is true for any natural number $n$.

Question 19: A sequence ${d_1},{d_2},{d_3}...$ is defined by letting ${d_1} = 2$ and ${d_k} = \dfrac{{{d_{k - 1}}}}{k}$ for all natural numbers $k \geqslant 2$. Show that ${d_n} = \dfrac{2}{{n!}}$ for all $n \in N$.

Ans: Given that ${d_n} = \dfrac{2}{{n!}}$

Put $n = 1$

$P(1):{d_1} = \dfrac{2}{{1!}}$

${d_1} = 2$

Put $n = 2$

$P(2):{d_2} = \dfrac{2}{{2!}}$

${d_2} = 1$

And ${d_k} = \dfrac{{{d_{k - 1}}}}{k}$

$\Rightarrow {d_2} = \dfrac{{{d_1}}}{2} = \dfrac{2}{2} = 1$

Thus, $P(2)$ is true.

Assume that $P(n)$ is true for a natural number $n = m$.

$P(m):{d_m} = \dfrac{2}{{m!}}$                                    $...(1)$

Now, we should prove that $P(m + 1)$ is true.

\[P(m + 1):{d_{m + 1}} = \dfrac{2}{{(m + 1)!}}\]

\[{d_{m + 1}} = \dfrac{{{d_{m + 1 - 1}}}}{{m + 1}}\]

\[{d_{m + 1}} = \dfrac{{{d_m}}}{{m + 1}}\]

From $(1)$

\[{d_{m + 1}} = \dfrac{2}{{m!(m + 1)}}\]

As we know that $n!(n + 1) = (n + 1)!$ ,

\[ \Rightarrow {d_{m + 1}} = \dfrac{2}{{(m + 1)!}}\]

Hence, $P(m + 1)$ is true whenever $P(m)$ is true.

So, by the principle of mathematical induction, 

$P(n)$ is true for any natural number $n$.

Question 20: Prove that for all 

$n \in N$$cos\alpha  + cos(\alpha  + \beta ) + cos(\alpha  + 2\beta ) + ... + cos(\alpha  + (n- 1)\beta ) = \dfrac{{cos\left( {\alpha  + \left( {\dfrac{{n - 1}}{2}} \right)\beta } \right)sin\left( {\dfrac{{n\beta }}{2}} \right)}}{{sin\dfrac{\beta }{2}}}$

Ans: Given that: $P(n):\cos \alpha  + \cos (\alpha  + \beta ) + \cos (\alpha  + 2\beta ) + ... + \cos (\alpha  + (n - 1)\beta ) = \dfrac{{\cos \left( {\alpha  + \left( {\dfrac{{n - 1}}{2}} \right)\beta } \right)\sin \left( {\dfrac{{n\beta }}{2}} \right)}}{{\sin \dfrac{\beta }{2}}}$

Put $n = 1$

$P(1):\cos \alpha  = \dfrac{{\cos \left( {\alpha  + \left( {\dfrac{{1 - 1}}{2}} \right)\beta } \right)\sin \left( {\dfrac{\beta }{2}} \right)}}{{\sin \dfrac{\beta }{2}}}$

$P(1):\dfrac{{\cos \alpha \sin \dfrac{\beta }{2}}}{{\sin \dfrac{\beta }{2}}}$

$P(1):\cos \alpha$

Hence, $P(1)$ is true.

Assume that $P(n)$ is true for a natural number $n = k$.

$P(k):\cos \alpha  + \cos (\alpha  + \beta ) + \cos (\alpha  + 2\beta ) + ... + \cos (\alpha  + (k - 1)\beta ) = \dfrac{{\cos \left( {\alpha  + \left( {\dfrac{{k - 1}}{2}} \right)\beta } \right)\sin \left( {\dfrac{{k\beta }}{2}} \right)}}{{\sin \dfrac{\beta }{2}}}{\text{   }}...(1)$Now, to prove that $P(k + 1)$ is true, we have to show that:

$P(k + 1):\cos \alpha  + \cos (\alpha  + \beta ) + \cos (\alpha  + 2\beta ) + ... + \cos (\alpha  + (k + 1 - 1)\beta ) = \dfrac{{\cos \left( {\alpha  + \left( {\dfrac{{k + 1 - 1}}{2}} \right)\beta } \right)\sin \left( {\dfrac{{(k + 1)\beta }}{2}} \right)}}{{\sin \dfrac{\beta }{2}}}$From $(1)$$\cos \alpha  + \cos (\alpha  + \beta ) + \cos (\alpha  + 2\beta ) + ... + \cos (\alpha  + (k - 1)\beta ) + \cos (\alpha  + k\beta ) = \dfrac{{\cos \left( {\alpha  + \left( {\dfrac{{k - 1}}{2}} \right)} \right)\sin \left( {\dfrac{{k\beta }}{2}} \right)}}{{\sin \dfrac{\beta }{2}}} + \cos (\alpha  + k\beta )$$P(n) = \dfrac{{\cos \left( {\alpha  + \left( {\dfrac{{k - 1}}{2}} \right)\beta } \right)\sin \left( {\dfrac{{k\beta }}{2}} \right) + \cos (\alpha  + k\beta )\sin \dfrac{\beta }{2}}}{{\sin \dfrac{\beta }{2}}}$

\[P(n) = \dfrac{{\sin \left( {\alpha  + \dfrac{{k\beta }}{2} - \dfrac{\beta }{2} + \dfrac{{k\beta }}{2}} \right) - \sin \left( {\alpha  + \dfrac{{k\beta }}{2} - \dfrac{\beta }{2} - \dfrac{{k\beta }}{2}} \right) + \sin \left( {\alpha  + k\beta  + \dfrac{\beta }{2}} \right) - \sin \left( {\alpha  + k\beta  - \dfrac{\beta }{2}} \right)}}{{2\sin \dfrac{\beta }{2}}}\]

\[P(n) = \dfrac{{\sin \left( {\alpha  + k\beta  + \dfrac{\beta }{2}} \right) - \sin \left( {\alpha  - \dfrac{\beta }{2}} \right)}}{{2\sin \dfrac{\beta }{2}}}\]

\[P(n) = \dfrac{{2\cos \dfrac{1}{2}\left( {\alpha  + k\beta  + \dfrac{\beta }{2} + \alpha  - \dfrac{\beta }{2}} \right)\sin \dfrac{1}{2}\left( {\alpha  + k\beta  + \dfrac{\beta }{2} - \alpha  + \dfrac{\beta }{2}} \right)}}{{2\sin \dfrac{\beta }{2}}}\]

\[P(n) = \dfrac{{\cos \left( {\dfrac{{2\alpha  + k\beta }}{2}} \right)\sin \left( {\dfrac{{k\beta  + \beta }}{2}} \right)}}{{\sin \dfrac{\beta }{2}}}\]

\[P(n) = \dfrac{{\cos \left( {\alpha  + \dfrac{{k\beta }}{2}} \right)\sin (k + 1)\dfrac{\beta }{2}}}{{\sin \dfrac{\beta }{2}}}\]

Hence, $P(k + 1)$ is true when $P(k)$ is true.

So, by the principle of mathematical induction,

$P(n)$ is true for any natural number $n$.

Question 21: Prove that, $cos\theta cos2\theta cos{2^2}\theta ...cos{2^{n - 1}}\theta  = \dfrac{{sin{2^n}\theta }}{{{2^n}sin\theta }}$, for all $n \in N$.

Ans: Given that $P(n):\cos \theta \cos 2\theta \cos {2^2}\theta ...\cos {2^{n - 1}}\theta  = \dfrac{{\sin {2^n}\theta }}{{{2^n}\sin \theta }}$

Put $n = 1$

$P(1):\dfrac{{\sin {2^1}\theta }}{{{2^1}\sin \theta }}$

$P(1):\dfrac{{2\sin \theta \cos \theta }}{{2\sin \theta }}$

$P(1):\cos \theta$ 

It is true.

Hence, $P(1)$ is true.

Assume that $P(n)$ is true for a natural number $n = k$.

$P(k):\cos \theta \cos 2\theta \cos {2^2}\theta ...\cos {2^{k - 1}}\theta  = \dfrac{{\sin {2^k}\theta }}{{{2^k}\sin \theta }}$                                    $...(1)$

Now, to prove that $P(k + 1)$ is true, we have to show that:

$P(k + 1):\cos \theta \cos 2\theta \cos {2^2}\theta ...\cos {2^{k - 1}}\theta \cos {2^k}\theta  = \dfrac{{\sin {2^{k + 1}}\theta }}{{{2^{k + 1}}\sin \theta }}$

From $(1)$

Now,$\cos \theta \cos 2\theta \cos {2^2}\theta ...\cos {2^{k - 1}}\theta \cos {2^k}\theta  = \dfrac{{\sin {2^k}\theta }}{{{2^k}\sin \theta }}\cos {2^k}\theta$

$P(k + 1):\dfrac{{2\sin {2^k}\theta \cos {2^k}\theta }}{{{{2.2}^k}\sin \theta }}$

$P(k + 1):\dfrac{{\sin {{2.2}^k}\theta }}{{{2^{k + 1}}\sin \theta }}$

$P(k + 1):\dfrac{{\sin {2^{(k + 1)}}\theta }}{{{2^{k + 1}}\sin \theta }}$

Hence, $P(k + 1)$ is true when $P(k)$ is true.

So, by the principle of mathematical induction,

$P(n)$ is true for any natural number $n$.

Question 22: Prove that, \[sin\theta  + sin2\theta  + sin3\theta  + ... + sinn\theta  = \dfrac{{sin\dfrac{{n\theta }}{2}sin\dfrac{{(n + 1)}}{2}\theta }}{{sin\dfrac{\theta }{2}}}\], for all $n \in N$.

Ans: Given that \[P(n):\sin \theta  + \sin 2\theta  + \sin 3\theta  + ... + \sin n\theta  = \dfrac{{\sin \dfrac{{n\theta }}{2}\sin \dfrac{{(n + 1)}}{2}\theta }}{{\sin \dfrac{\theta }{2}}}\]

Put $n = 1$

$P(1):\sin \theta  = \dfrac{{\sin \dfrac{\theta }{2}.\sin \dfrac{{(1 + 1)}}{2}\theta }}{{\sin \dfrac{\theta }{2}}}$

$P(1):\dfrac{{\sin \dfrac{\theta }{2}.\sin \theta }}{{\sin \dfrac{\theta }{2}}}$

$P(1):\sin \theta$ 

Hence, $P(1)$ is true.

Assume that $P(n)$ is true for a natural number $n = k$.

\[P(k):\sin \theta  + \sin 2\theta  + \sin 3\theta  + ... + \sin k\theta  = \dfrac{{\sin \dfrac{{k\theta }}{2}\sin \left( {\dfrac{{k + 1}}{2}} \right)\theta }}{{\sin \dfrac{\theta }{2}}}\]                                    $...(1)$

Now, to prove that $P(k + 1)$ is true, we have to show that:

\[P(k + 1):\sin \theta  + \sin 2\theta  + \sin 3\theta  + ... + \sin k\theta  + \sin (k + 1)\theta  = \dfrac{{\sin \dfrac{{(k + 1)\theta }}{2}\sin \left( {\dfrac{{k + 1 + 1}}{2}} \right)\theta }}{{\sin \dfrac{\theta }{2}}}\]

From $(1)$

\[P(k + 1):\sin \theta  + \sin 2\theta  + \sin 3\theta  + ... + \sin k\theta  + \sin (k + 1)\theta  = \dfrac{{\sin \dfrac{{k\theta }}{2}\sin \left( {\dfrac{{k + 1}}{2}} \right)\theta }}{{\sin \dfrac{\theta }{2}}} + \sin (k + 1)\theta \]

\[P(k + 1):\dfrac{{\sin \dfrac{{k\theta }}{2}\sin \left( {\dfrac{{k + 1}}{2}} \right)\theta  + \sin (k + 1)\theta .\sin \dfrac{\theta }{2}}}{{\sin \dfrac{\theta }{2}}}\]

\[P(k + 1):\dfrac{{\cos \left[ {\dfrac{{k\theta }}{2} - \left( {\dfrac{{k + 1}}{2}} \right)\theta } \right] - \cos \left[ {\dfrac{{k\theta }}{2} + \left( {\dfrac{{k + 1}}{2}} \right)\theta } \right] + \cos \left[ {(k + 1)\theta  - \dfrac{\theta }{2}} \right] - \cos \left[ {(k + 1)\theta  + \dfrac{\theta }{2}} \right]}}{{2\sin \dfrac{\theta }{2}}}\]

\[P(k + 1):\dfrac{{\cos \dfrac{\theta }{2} - \cos \left( {k\theta  + \dfrac{\theta }{2}} \right) + \cos \left( {k\theta  + \dfrac{\theta }{2}} \right) - \cos \left( {k\theta  + \dfrac{{3\theta }}{2}} \right)}}{{2\sin \dfrac{\theta }{2}}}\]

\[P(k + 1):\dfrac{{\cos \dfrac{\theta }{2} - \cos \left( {k\theta  + \dfrac{{3\theta }}{2}} \right)}}{{2\sin \dfrac{\theta }{2}}}\]

\[P(k + 1):\dfrac{{2\sin \dfrac{1}{2}\left( {\dfrac{\theta }{2} + k\theta  + \dfrac{{3\theta }}{2}} \right).\sin \dfrac{1}{2}\left( {k\theta  + \dfrac{{3\theta }}{2} - \dfrac{\theta }{2}} \right)}}{{2\sin \dfrac{\theta }{2}}}\]

\[P(k + 1):\dfrac{{\sin \left( {\dfrac{{k\theta  + 2\theta }}{2}} \right).\sin \left( {\dfrac{{k\theta  + \theta }}{2}} \right)}}{{\sin \dfrac{\theta }{2}}}\]

\[P(k + 1):\dfrac{{\sin (k + 1)\dfrac{\theta }{2}.\sin (k + 1 + 1)\dfrac{\theta }{2}}}{{\sin \dfrac{\theta }{2}}}\]

Hence, $P(k + 1)$ is true when $P(k)$ is true.

So, by the principle of mathematical induction,

$P(n)$ is true for any natural number $n$.

Question 23: Show that $\dfrac{{{n^5}}}{5} + \dfrac{{{n^3}}}{3} + \dfrac{{7n}}{{15}}$ is a natural number for all $n \in N$.

Ans: Given that $P(n):\dfrac{{{n^5}}}{5} + \dfrac{{{n^3}}}{3} + \dfrac{{7n}}{{15}}$

Put $n = 1$

$P(1):\dfrac{{{1^5}}}{5} + \dfrac{{{1^3}}}{3} + \dfrac{7}{{15}}$

$P(1):\dfrac{{15}}{{15}}$

$P(1):1$ 

Hence, $P(1)$ is true.

Assume that $P(n)$ is true for a natural number $n = k$.

$P(k):\dfrac{{{k^5}}}{5} + \dfrac{{{k^3}}}{3} + \dfrac{{7k}}{{15}}$ is a natural number.                                    $...(1)$

Now, we should prove that $P(k + 1)$ is true.

$P(k + 1):\dfrac{{{{(k + 1)}^5}}}{5} + \dfrac{{{{(k + 1)}^3}}}{3} + \dfrac{{7(k + 1)}}{{15}}$

$P(k + 1):\dfrac{{{k^5} + 5{k^4} + 10{k^3} + 10{k^2} + 5k + 1}}{5} + \dfrac{{{k^3} + 1 + 3{k^2} + 3k}}{3} + \dfrac{{7k + 7}}{{15}}$

$P(k + 1):\dfrac{{{k^5}}}{5} + \dfrac{{{k^3}}}{3} + \dfrac{{7k}}{{15}} + \dfrac{{5{k^4} + 10{k^3} + 10{k^2} + 5k + 1}}{5} + \dfrac{{1 + 3{k^2} + 3k}}{3} + \dfrac{7}{{15}}$

$P(k + 1):\dfrac{{{k^5}}}{5} + \dfrac{{{k^3}}}{3} + \dfrac{{7k}}{{15}} + {k^4} + 2{k^3} + 2{k^2} + k + {k^2} + k + \dfrac{1}{5} + \dfrac{1}{3} + \dfrac{7}{{15}}$

From $(1)$

$P(k + 1):\dfrac{{{k^5}}}{5} + \dfrac{{{k^3}}}{3} + \dfrac{{7k}}{{15}} + {k^4} + 2{k^3} + 3{k^2} + 2k + 1$ is a natural number.

Hence, $P(k + 1)$ is true whenever $P(k)$ is true.

So, by the principle of mathematical induction,

$P(n)$ is true for any natural number $n$.

Question 24: Prove that $\dfrac{1}{{n + 1}} + \dfrac{1}{{n + 2}} + ... + \dfrac{1}{{2n}} > \dfrac{{13}}{{24}}$ for all natural numbers n>1.

Ans: Given that $P(n):\dfrac{1}{{n + 1}} + \dfrac{1}{{n + 2}} + ... + \dfrac{1}{{2n}} > \dfrac{{13}}{{24}}$

Put $n = 2$

$P(2):\dfrac{1}{2} + \dfrac{1}{4} = \dfrac{3}{4} > \dfrac{{13}}{{24}}$ is true

$P(3):\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} = \dfrac{{11}}{{12}} > \dfrac{{13}}{{24}}$ is true

Assume that $P(n)$ is true for a natural number $n = k$.

$P(k):\dfrac{1}{{k + 1}} + \dfrac{1}{{k + 2}} + ... + \dfrac{1}{{2k}} > \dfrac{{13}}{{24}}$ 

Now, we should prove that $P(k + 1)$ is true.

$P(k + 1):\dfrac{1}{{(k + 1) + 1}} + \dfrac{1}{{(k + 1) + 2}} + ... + \dfrac{1}{{2(k + 1)}}$

$P(k + 1):\dfrac{1}{{k + 2}} + \dfrac{1}{{k + 3}} + ... + \dfrac{1}{{2k + 2}}$

$P(k + 1):\dfrac{1}{{k + 1}} + \dfrac{1}{{k + 2}} + \dfrac{1}{{k + 3}} + ... + \dfrac{1}{{2k}} + \dfrac{1}{{2k + 1}} + \dfrac{1}{{2k + 2}} - \dfrac{1}{{k + 1}}$

$P(k + 1):\dfrac{1}{{k + 1}} + \dfrac{1}{{k + 2}} + \dfrac{1}{{k + 3}} + ... + \dfrac{1}{{2k}} + \dfrac{1}{{2k + 1}} - \dfrac{1}{{2k + 2}}$

$P(k + 1):\dfrac{1}{{k + 1}} + \dfrac{1}{{k + 2}} + \dfrac{1}{{k + 3}} + ... + \dfrac{1}{{2k}} + \dfrac{1}{{(2k + 1)(2k + 2)}} > \dfrac{{13}}{{24}}$

Hence, $P(k + 1)$ is true whenever  $P(k)$ is true.

So, by the principle of mathematical induction,

$P(n)$ is true for natural numbers $n > 1$.

Question 25: Prove that number of subsets of a set containing n distinct elements is ${2^n}$, for all $n \in N$.

Ans: $P(n):$ Number of subset of a sethaving $n$ distinct elements is ${2^n}$ for all $n \in N$

Take $n = 1$,

Consider a set with one element:

 $A = \{ 1\}$

The set of subsets of the above set is \[\{ \{ 1\} ,\phi \} \] and it contains ${2^1}$ elements.

$P(1)$ is true.

Assume $P(n)$ is true for a natural number $n = k$.

$p(k):$ Number of subsets of a set having $k$ distinct elements is ${2^k}$.

We should prove that $p(k + 1)$is true. 

$p(k + 1):$ Number of subsets of a set having $(k + 1)$ distinct elements is ${2^{k + 1}}$.

The number of subsets is usually doubled whenever an element is added to the given set.

Number of subsets containing $(k + 1)$ distinct elements is: $2 \times {2^k} = {2^{k + 1}}$.

Thus, we can say that $P(k + 1)$ is true.

Therefore, the number of subsets of a set containing $n$  distinct elements is  ${2^n}$ for all  $n \in N$.

Question 26: If $1{0^n} + 3.{4^{n + 2}} + k$ is divisible by 9 for all $n \in N$, then the      least positive integer value of k is:

(a) 5

(b) 3

(c) 7

(d) 1

Ans: Option a is correct.

Given that $P(n):{10^n} + {3.4^{n + 2}} + k$

Put $n = 1$

$P(1):{10^1} + {3.4^{1 + 2}} + k$

$P(1):2012 + k$ 

As it is exactly divisible by $9$, the value of $k$ will be $5$.

So, option A is the correct answer.

Question 27: For all $n \in N$, $3.{5^{2n + 1}} + {2^{3n + 1}}$ is divisible by 

(a) 19

(b) 17

(c) 23

(d) 25

Ans: Option b is correct.

Given that $P(n):{3.5^{2n + 1}} + {2^{3n + 1}}$

Put $n = 1$

$P(1):{3.5^{2 + 1}} + {2^{3 + 1}}$

$P(1):375 + 16$

$P(1):391$

$P(1):17 \times 23$

Put $n = 2$

$P(2):{3.5^{4 + 1}} + {2^{6 + 1}}$

$P(2):9375 + 256$

$P(2):9503$

$P(2):17 \times 559$ which is divisible by $17$.

Hence, for all $n \in N$, ${3.5^{2n + 1}} + {2^{3n + 1}}$ is divisible by $17$.

Question 28: If ${x^n} - 1$ is divisible by $x - k$, then the least positive integer value of k is

(a) 1

(b) 2

(c) 3

(d) 4

Ans: Option a is correct. 

Given that $P(n):{x^n} - 1 = {\lambda _n}(x - k)$

Put $n = 1$

$P(1):x - 1 = {\lambda _1}(x - k)$

Put $n = 2$

$P(2):{x^2} - 1 = {\lambda _2}(x - k)$

$P(2):(x - 1)(x + 1) = {\lambda _2}(x - k)$

So, the least positive integral value of $k$ is $k = 1$.

Question 29: If $P(n):2n < n!,\;n \in N$ then P(n) is true for all $n \geqslant \_\_\_\_\_.$

Ans: Let $P(n):2n < n!,n \in N$

Put $n = 1$

$P(1):2 < 1$ which is false.

Put $n = 2$

$P(2):4 < 2$ which is false.

Put $n = 3$

$P(3):6 < 6$ which is false.

Put $n = 4$

$P(4):8 < 24$ which is true.

Hence, $P(n)$ is true for all $n \geqslant 4$.

Question 30: Let P(n) be a statement and let \[P(k) \Rightarrow P(k + 1)\], for some natural number k then P(n) is true for all $n \in N$.

Ans: The given statement is true.

We have $P(k) \Rightarrow P(k + 1)$

Substitute $k - 1$ in place of $k$.

$P(k - 1) \Rightarrow P(k)$

So, if $P(k)$ is true for some $n \in N$ then it is true for $(k - 1)$ and $(k + 1)$.

Hence, it is true for all $n \in N$.

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