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Binomial Theorem Class 11 Notes Maths - Basic Subjective Questions
Section–A (1 Mark Questions)
1. Find the number of terms in the expansion of (1+x)10 + (1-x)10.
Ans. The expansion of $(1+x)^{10}+(1-x)^{10}$ is given as
$\begin{aligned} & {\left[{ }^{10} C_0 x^0+{ }^{10} C_1 x^1+{ }^{10} C_2 x^2+\ldots+{ }^{10} C_{10} x^{10}\right]+} \\ & {\left[{ }^{10} C_0(-x)^0+{ }^{10} C_1(-x)^1+{ }^{10} C_2(-x)^2+\ldots+{ }^{10} C_{10}(-x)^{10}\right]}\end{aligned}$
$=2\left[{ }^{10} C_0 x^0+{ }^{10} C_2 x^2+{ }^{10} C_4 x^4+\ldots+{ }^{10} C_{10} x^{10}\right]$
Hence, the number of terms in the given expansion is 6 .
2. What is the value of $\sum_{r=0}^{r=n} 4^{r n} C_r$ .
Ans.
$$ \begin{aligned} & \sum_{r=0}^{r=n} 4^{r n} C_r=4^{0 . n} C_0+4^{1 .} C_1+4^{2 n} \cdot C_2+\ldots+4^{n . n} C_n \\ & \quad={ }^n C_0 \cdot 4^0+{ }^n C_1 \cdot 4^1+{ }^n C_2 \cdot 4^2+\ldots+{ }^n C_n \cdot 4^n \\ & =(1+4)^n=5^n . \end{aligned} $$
3. Find the general term in the expansion of $\left(2 a+\frac{b}{2}\right)^n$,$n\epsilon N$.
Ans. The general term in the expansion of $\left(2 a+\frac{b}{2}\right)^n$ is -
$$T_{r+1}={ }^n C_r \cdot(2 a)^{n-r}\left(\frac{b}{2}\right)^r=2^{n-2 r n} \cdot C_r \cdota^{n-r} b^r$$
4. Using binomial theorem, write down the expansion of (1-3x)7 .
Ans. The binomial expansion of $(1-3 x)^7$ is given by
$$ \begin{aligned} & { }^7 C_4(3 x)^6-{ }^7 C_1(3 x)^1+{ }^7 C_2(3 x)^2-{ }^7 C_1(3 x)^3+{ }^7 C_4(3 x)^4 \\ & -{ }^7 C_5(3 x)^5+{ }^7 C_6(3 x)^6-{ }^7 C_7(3 x)^7 \\ & =1-21 x+189 x^2-945 x^3+2835 x^4 \\ & -5103 x^5+5103 x^6-2187 x^7 . \end{aligned} $$
5. Find the 5th term from the end in the expansion of $\left(3 x-\frac{1}{x^2}\right)^{10}$.
Ans. The $5^{\text {th }}$ term from the end in $\left(3 x-\frac{1}{x^2}\right)^{10}=5^{\text {th }}$ term from the beginning in $\left(\frac{1}{x^2}-3 x\right)^{10}$ which is given by
$$ \begin{aligned} T_5=T_{4+1} & ={ }^{10} C_4\left(\frac{1}{x^2}\right)^{10-4}(-3 x)^4 \\ & =3^{4.10} C_4 x^{-8} \\ & =\frac{10 \times 9 \times 8 \times 7 \times 81}{4 \times 3 \times 2 \times 1 \times x^8} \\ & =\frac{17010}{x^8} \end{aligned} $$
Section–B (2 Marks Questions)
6. Find the 8th term in the expansion of $\left(x^{\frac{3}{2}} y^{\frac{1}{2}}-x^{\frac{1}{2}} y^{\frac{3}{2}}\right)^{10}$ .
Ans. The $8^{\text {sh }}$ term in the expansion of $\left(x^{\frac{3}{2}} y^{\frac{1}{2}}-x^{\frac{1}{2}} y^{\frac{3}{2}}\right)^{10}$ is given by-
$$ \begin{aligned} \because \quad T_8 & =T_{7+1} \\ T_8=T_{7+1} & ={ }^{10} C_7\left(x^{\frac{3}{2}} y^{\frac{1}{2}}\right)^{10-7}\left(-x^{\frac{1}{2}} y^{\frac{3}{2}}\right)^7 \\ & =-\frac{10 !}{7 ! \times 3 !} x^{\frac{9}{2}+\frac{7}{2}} \cdot y^{\frac{3}{2}+\frac{21}{2}} \\ & =-\frac{10 \times 9 \times 8}{3 \times 2} x^8 y^{12} \\ & =-120 x^8 y^{12} \end{aligned} $$
7. If p is a real number and if the middle term in the expansion of $\left(\frac{p}{2}+2\right)^8$ is 1120, then find the value of p.
Ans. In the binomial expansion of $\left(\frac{p}{2}+2\right)^8$, we observe that $\left(\frac{8}{2}+1\right)^{\text {sh }}$ i.e., $5^{\text {th }}$ term is the middle term.
It is given that the middle term is 1120 .
$$ \begin{aligned} & \therefore T_5=1120 \\ & \Rightarrow{ }^8 C_4\left(\frac{p}{2}\right)^{8-4}(2)^4=1120 \\ & \Rightarrow \frac{8 !}{4 ! \times 4 !} p^4=1120 \\ & \Rightarrow p^4=16 \\ & \Rightarrow p= \pm 2 \end{aligned} $$
Hence, the real values of $p$ is \pm 2 .
8. Find the Constant term in the expansion of $\left(x-\frac{1}{x}\right)^{10}$ .
Ans. The general term for the expansion of $\left(x-\frac{1}{x}\right)^{10}$ can be given by
$$ \begin{aligned} T_{r+1} & ={ }^{10} C_r(x)^{10-r}\left(-\frac{1}{x}\right)^r \\ & =(-1)^r{ }^{10} C_r(x)^{10-2 r} \end{aligned} $$
For constant term,
$$\begin{aligned}& 10-2 r=0 \\& \Rightarrow r=5\end{aligned}$$
Therefore,
$$T_{5+1}={ }^{10} C_5(-1)^5=-252$$
Hence, the constant term is -252 .
9. Does the expansion $\left(2 x^2-\frac{1}{x}\right)^{20}$ contain any term involving x9.
Ans. Suppose $x^9$ occurs in the given expression $\left(2 x^2-\frac{1}{x}\right)^{20}$ at the $(r+1)^{t h}$ term.
Then, we have
$$T_{r+1}={ }^{20} C_r\left(2 x^2\right)^{20-r}\left(-\frac{1}{x}\right)^r$$
$$=(-1)^r \cdot{ }^{20} C_r(2)^{20-r}(x)^{10-2 r-r}$$
For this term to contain $x^9$, we must have
$$\begin{aligned}& 40-3 r=9 \\& \Rightarrow 3 r=31 \\& \Rightarrow r=\frac{31}{3}\end{aligned}$$
It is not possible, as $r$ is not an integer.
Hence, there is no term with $x^9$ in the given expression.
10. Prove that the coefficient of $(r+1)^{\text {th }}$ term in the expansion of $(1+x)^{n+1}$ is equal to the sum of the coefficients of rth and (r+1)th terms in the expansion of (1+x)n.
Ans. Coefficient of the $(r+1)^{\text {th }}$ term in $(1+x)^{n+1}$ is ${ }^{n+1} C_{\text {. }}$.
Coefficient of the $r^{\text {th }}$ term in $(1+x)^n$ is ${ }^n C_{r-1}$.
Coefficient of the $(r+1)^{16}$ term in $(1+x)^n$ is " $C_r$.
Therefore,
Sum of the coefficients of the $r^{\text {tk }}$ and $(r+1)^{t h}$ terms in $(1+x)^n={ }^n C_{r-1}+{ }^n C_r$ $={ }^{n+1} C_r \quad\left[\because{ }^a C_{r-1}+{ }^n C_r={ }^{n+1} C_r\right]$
Hence proved.
11. Find the 4th term from the end in the expansion of $\left(\frac{4 x}{5}-\frac{5}{2 x}\right)^8$ .
Ans. Let $T_4$ be the $4^{\text {tk }}$ term from the end of the given expression, Then,
$4^{\text {th }}$ term from the end in $\left(\frac{4 x}{5}-\frac{5}{2 x}\right)^8=4^{\text {in }}$ term from the beginning in $\left(\frac{5}{2 x}-\frac{4 x}{5}\right)^8$ Thus, we have
$$T_4=T_{3+1}={ }^8 C_3\left(\frac{5}{2 x}\right)^{8-3}\left(-\frac{4 x}{5}\right)^3$
$$\begin{aligned}& =-{ }^3 C_3\left(\frac{5}{2}\right)^5\left(\frac{4}{5}\right)^3 x^{-2} \\& =-C_3 \cdot 5^2 \cdot 2^1 x^{-2} \\& =-\frac{8 !}{3 ! \times 5 !} \cdot \frac{50}{x^2} \\& =-\frac{8.7 \cdot 6}{3.2} \cdot \frac{50}{x^2} \\& =-\frac{2800}{x^2} .\end{aligned}$$
12. Find the term independent of x in the expansion of $\left(2 x+\frac{1}{3 x^2}\right)^9$ .
Ans. Suppose the $(r+1)^{t h}$ term in the given expansion $\left(2 x+\frac{1}{3 x^2}\right)^9$ is independent of $x$.
Thus,
$$T_{r+1}={ }^9 C_r(2 x)^{9-r}\left(\frac{1}{3 x^2}\right)^r={ }^9 C_r \cdot \frac{2^{9-r}}{3^r} x^{9-3 r}$$
For this term to be independent of $x$, we must have
$$\begin{aligned}& 9-3 r=0 \\& \Rightarrow r=3\end{aligned}$$
Hence, the required term is the $4^{\text {tit }}$ term which can be written as
$$\begin{aligned}T_4=T_{3+1} & ={ }^9 C_3 \frac{2^{9-3}}{3^3}={ }^9 C_3 \times \frac{2^6}{3^3} \\& =\frac{9 !}{3 ! 6 !} \times \frac{2^6}{3^3}=\frac{1792}{9} .\end{aligned}$$
13. If the coefficients of $(2r+1)^{th}$ term and $(r+2)^{th}$ term in the expansion of (1+x)^{43}are equal, find r.
Ans. We know that the coefficient of the $r^{\text {th }}$ term in the expansion of $(1+x)^n$ is " $C_{r-1}$.
Therefore, the coefficients of the $(2 r+1)^{k r}$ and $(r+2)^{\text {th }}$ terms in the given expression are ${ }^{43} C_{2 r}$ and ${ }^{43} C_{r \rightarrow 1}$.
For these coefficients to be equal, we must have,
2r= r+1 or 2r + r + 1 = 43
$\left [ Q\;^{n}C_{r}=^{n}C_{s} \Rightarrow r=s\; or\;r+s=n\right ]$
$\Rightarrow$ r=1 or r=14
$\Rightarrow$ r=14
($\because$ for r=1 it gives the same term).
PDF Summary - Class 11 Maths Binomial Theorem Notes (Chapter 7)
The computations become harder when employing repeated multiplication for higher powers like ${{\left( 93 \right)}^{8}},{{\left( 105 \right)}^{7}},{{\left( 305 \right)}^{4}},...$ and so on.
A theorem known as the binomial theorem was used to solve this problem.
It simplifies the expansion of ${{\left( \text{a+b} \right)}^{\text{n}}}$, where $\text{n}$ is an integer or a rational number.
Binomial Theorem
If \[\text{a, b }\in \text{ R}\]and \[\text{n }\in \text{ N}\] , then
\[{{\left( \text{a+b} \right)}^{\text{n}}}\text{=}{}^{\text{n}}{{\text{C}}_{\text{0}}}{{\text{a}}^{\text{n}}}{{\text{b}}^{\text{0}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{1}}}{{\text{a}}^{\text{n-1}}}{{\text{b}}^{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}{{\text{a}}^{\text{n-2}}}{{\text{b}}^{\text{2}}}\text{+}.....\text{+}{}^{\text{n}}{{\text{C}}_{\text{n}}}{{\text{a}}^{\text{0}}}{{\text{b}}^{\text{n}}}\]
Remarks :
If the binomial's index is $\text{n}$, the expansion will have $\text{n+1}$ terms .
The total sum of the indices of $\text{a}$ and $\text{b}$ in each term is always $\text{n}$.
In a binomial expansion equidistant from both ends, the coefficients of the terms are equal.
\[{{\left( \text{a-b} \right)}^{\text{n}}}\text{=}{}^{\text{n}}{{\text{C}}_{\text{0}}}{{\text{a}}^{\text{n}}}{{\text{b}}^{\text{0}}}-{}^{\text{n}}{{\text{C}}_{\text{1}}}{{\text{a}}^{\text{n-1}}}{{\text{b}}^{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}{{\text{a}}^{\text{n-2}}}{{\text{b}}^{\text{2}}}\text{+}.....\text{+}\left( -1 \right){}^{\text{n}}{{\text{C}}_{\text{n}}}{{\text{a}}^{\text{0}}}{{\text{b}}^{\text{n}}}\]
Binomial Theorem for Positive Integral Indices
\[{{\left( \text{a+b} \right)}^{\text{0}}}\text{=1}\], where \[\text{a+b }\ne 0\]
\[{{\left( \text{a+b} \right)}^{\text{1}}}\text{= a+b}\]
\[{{\left( \text{a+b} \right)}^{\text{2}}}\text{= }{{\text{a}}^{\text{2}}}\text{+2ab+}{{\text{b}}^{\text{2}}}\]
\[{{\left( \text{a+b} \right)}^{\text{3}}}\text{= }{{\text{a}}^{\text{3}}}\text{+3}{{\text{a}}^{\text{2}}}\text{b+3a}{{\text{b}}^{\text{2}}}\text{+}{{\text{b}}^{\text{3}}}\]
\[{{\left( \text{a+b} \right)}^{\text{4}}}\text{=}{{\left( \text{a+b} \right)}^{\text{3}}}\text{. }\left( \text{a+b} \right)\text{=}{{\text{a}}^{\text{4}}}\text{+4}{{\text{a}}^{\text{3}}}\text{b+6}{{\text{a}}^{\text{2}}}{{\text{b}}^{\text{2}}}\text{+4a}{{\text{b}}^{\text{3}}}\text{+}{{\text{b}}^{\text{4}}}\]
We can see in these expansions that
There are one more terms in the expansion than the terms present in the index.
In each term, the powers of the first quantity ‘$\text{a}$' decrease by one, while the powers of the second quantity ‘$\text{b}$' increase by one.
The index of $\left( \text{a+b} \right)$ is equal to the total of the indices of $\text{a}$ and $\text{b}$ in each term of the expansion.
Pascal’s Triangle:
The expansion coefficients are organised in a triangle-shaped array with $\left( \text{1} \right)$ at the top vertex and running down the two slanting sides. This arrangements is known as “Pascal's triangle”.
Pingla also refers to it as Meru Prastara.
Pascal's triangle can also be used to expand binomials to their higher powers.
Here, Combination method is used.
\[{}^{\text{n}}{{\text{C}}_{\text{r}}}\text{=}\dfrac{\text{n!}}{\text{r!}\left( \text{n-r} \right)\text{!}}\] , \[\text{0}\le \text{r}\le \text{n}\]
Where, \[\text{n}\]is a non-negative number.
\[{}^{\text{n}}{{\text{C}}_{\text{0}}}\text{=}{}^{\text{n}}{{\text{C}}_{\text{n}}}\text{=1}\]
Binomial Theorem for any Positive Integer\[n\]:
\[{{\left( \text{a+b} \right)}^{\text{n}}}\text{=}{}^{\text{n}}{{\text{C}}_{\text{0}}}{{\text{a}}^{\text{n}}}{{\text{b}}^{\text{0}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{1}}}{{\text{a}}^{\text{n-1}}}{{\text{b}}^{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}{{\text{a}}^{\text{n-2}}}{{\text{b}}^{\text{2}}}\text{+}.....\text{+}{}^{\text{n}}{{\text{C}}_{\text{n}}}{{\text{a}}^{\text{0}}}{{\text{b}}^{\text{n}}}\]
That is,
\[{{\left( \text{a+b} \right)}^{\text{n}}}\text{=}{}^{\text{n}}{{\text{C}}_{\text{0}}}{{\text{a}}^{\text{n}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{1}}}{{\text{a}}^{\text{n-1}}}{{\text{b}}^{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}{{\text{a}}^{\text{n-2}}}{{\text{b}}^{\text{2}}}\text{+}.....\text{+}{}^{\text{n}}{{\text{C}}_{\text{n}}}{{\text{b}}^{\text{n}}}\]
Remarks:
Binomial theorem can also be written as,
\[{{\left( \text{a+b} \right)}^{\text{n}}}=\sum\limits_{\text{k=0}}^{\text{n}}{{}^{\text{n}}{{\text{C}}_{\text{k}}}{{\text{a}}^{\text{n-k}}}{{\text{b}}^{\text{k}}}}\]
Where, \[\sum\limits_{\text{k=0}}^{\text{n}}{{}^{\text{n}}{{\text{C}}_{\text{k}}}{{\text{a}}^{\text{n-k}}}{{\text{b}}^{\text{k}}}}\] represents
\[{}^{\text{n}}{{\text{C}}_{\text{0}}}{{\text{a}}^{\text{n}}}{{\text{b}}^{\text{0}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{1}}}{{\text{a}}^{\text{n-1}}}{{\text{b}}^{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}{{\text{a}}^{\text{n-2}}}{{\text{b}}^{\text{2}}}\text{+}.....\text{+}{}^{\text{n}}{{\text{C}}_{\text{n}}}{{\text{a}}^{\text{0}}}{{\text{b}}^{\text{n}}}\]
The coefficients \[{}^{\text{n}}{{\text{C}}_{\text{r}}}\]are known as binomial coefficients.
Example: Compute \[{{\left( 99 \right)}^{5}}\]
Ans:
We can consider \[99\] as follows,
\[99=100-1\]
Therefore,
\[{{\left( 99 \right)}^{5}}={{\left( 100-1 \right)}^{5}}\]
By applying Binomial Theorem for above, we get
${{\left( 99 \right)}^{5}}={}^{5}{{\text{C}}_{0}}{{\left( 100 \right)}^{5}}{{.1}^{0}}-{}^{5}{{\text{C}}_{1}}{{\left( 100 \right)}^{4}}{{.1}^{1}}+{}^{5}{{\text{C}}_{2}}{{\left( 100 \right)}^{3}}{{.1}^{2}}-{}^{5}{{\text{C}}_{3}}{{\left( 100 \right)}^{2}}{{.1}^{3}} +{}^{5}{{\text{C}}_{4}}{{\left( 100 \right)}^{1}}{{.1}^{4}}+{}^{5}{{\text{C}}_{5}}{{\left( 100 \right)}^{0}}{{.1}^{5}}$
$ =100000000005\times 100000000\times 1+10\times 1000000\times 110\times 10000\times 1 +5\times 100\times 11 $
\[=10000000000-500000000+10000000-100000+500-1\]
\[=9509900499\]
Therefore,
\[{{\left( 99 \right)}^{5}}=9509900499\]
General Term and Middle Terms in Expansion of \[{{\left( \text{a+b} \right)}^{\text{n}}}\]:
\[{{\text{t}}_{\text{r+1}}}\text{= }{{\text{ }}^{\text{n}}}{{\text{C}}_{\text{r}}}{{\text{a}}^{\text{n-r}}}+{{\text{b}}^{\text{r}}}\]
\[{{\text{t}}_{\text{r+1}}}\] is known as a general term for all \[\text{r}\in \text{N}\] and \[\text{0}\le \text{r}\le \text{n}\].
By using this formula, any term of the expansion can be calculated.
MIDDLE TERM \[\left( \text{S} \right)\] :
In \[{{\left( \text{a+b} \right)}^{\text{n}}}\] if \[\text{n}\] is even then the number of terms in the expansion is odd. Therefore there is only one middle term and it is \[{{\left( \dfrac{\text{n}+2}{2} \right)}^{\text{th}}}\] term.
In \[{{\left( \text{a+b} \right)}^{\text{n}}}\], if n is odd then the number of terms in the expansion is even. Therefore there are two middle terms and those are \[{{\left( \dfrac{\text{n}+1}{2} \right)}^{\text{th}}}\] and \[{{\left( \dfrac{\text{n}+3}{2} \right)}^{\text{th}}}\] terms.
Binomial Theorem for any Index:
If \[\text{n}\] is negative integer then \[\text{n!}\] cannot be defined. We state binomial theorem in another form.
\[{{\left( \text{a+b} \right)}^{\text{n}}}\text{=}{{\text{a}}^{\text{n}}}\text{+}\dfrac{\text{n}}{1!}{{\text{a}}^{\text{n-1}}}{{\text{b}}^{\text{1}}}\text{+}\dfrac{\text{n}\left( \text{n}-1 \right)}{2!}{{\text{a}}^{\text{n-2}}}{{\text{b}}^{\text{2}}}\text{+}.....\text{+}\dfrac{\text{n}\left( \text{n}-1 \right)\left( \text{n}-\text{r}+1 \right)}{\text{r}!}{{\text{a}}^{\text{n-r}}}{{\text{b}}^{\text{r}}}+...\]
Here,
\[{{\text{t}}_{\text{r+1}}}\text{ = }\dfrac{\text{n}\left( \text{n}-1 \right)\left( \text{n}-\text{r}+1 \right)}{\text{r}!}{{\text{a}}^{\text{n-r}}}{{\text{b}}^{\text{r}}}\]
Theorem :
If \[\text{n}\] is any real number,
\[\text{a = 1,b = x}\] and \[\left| \text{x} \right| < 1\] then
\[{{\left( 1+\text{x} \right)}^{\text{n}}}\text{= 1+ nx + }\dfrac{\text{n}\left( \text{n}-1 \right)}{\text{2}!}{{\text{x}}^{2}}+\text{ }\dfrac{\text{n}\left( \text{n}-1 \right)\left( \text{n}-2 \right)}{\text{3}!}{{\text{x}}^{3}}+...\]
Here there are infinite number of terms in the expansion, the general term is given by
\[{{\text{t}}_{\text{r+1}}}\text{ = }\dfrac{\text{n}\left( \text{n}-1 \right)\left( \text{n}-2 \right)\left( \text{n}-\text{r}+1 \right)}{\text{r}!}\text{, r}\ge \text{0}\]
Note:
Expansion is valid only when \[\text{-1 x 1}\]
\[^{\text{n}}{{\text{C}}_{\text{r}}}\] can not be used because it is defined only for natural number, so \[^{\text{n}}{{\text{C}}_{\text{r}}}\] can be written as \[\dfrac{\text{n}\left( \text{n}-1 \right)\left( \text{n}-2 \right)\left( \text{n}-\text{r}+1 \right)}{\text{r}!}\]
As the series never terminates, the number of terms in the series is infinite.
General term of the series \[{{\text{(1+x)}}^{-\text{n}}}={{\text{T}}_{\text{r+1}}}\to {{\left( -1 \right)}^{\text{r}}}\]
\[\dfrac{\text{1+x}}{\text{1-x}}\] if \[\left| \text{x} \right| < 1\]
General term of the series \[{{\text{(1+x)}}^{-\text{n}}}\to {{\text{T}}_{\text{r+1}}}\]
\[\text{= }\dfrac{\text{ }\left( \text{ -1} \right)\left( \text{ +2} \right)...\left( \text{ + -1} \right)}{\text{r!}}\text{x}\]
If first term is not $1$, then make it unity in the following way.
\[{{\text{(a+x)}}^{\text{n}}}={{\text{a}}^{\text{n}}}{{\left( \dfrac{\text{1+x}}{\text{a}} \right)}^{\text{n}}}\]if \[\left| \dfrac{\text{x}}{\text{a}} \right| < 1\]
Remarks:
If \[\left| \text{x} \right| < 1\] and \[\text{n}\]is any real number, then
\[{{\left( 1-\text{x} \right)}^{\text{n}}}\text{= 1- nx + }\dfrac{\text{n}\left( \text{n}-1 \right)}{\text{2}!}{{\text{x}}^{2}}-\text{ }\dfrac{\text{n}\left( \text{n}-1 \right)\left( \text{n}-2 \right)}{\text{3}!}{{\text{x}}^{3}}+...\]
The general term is given by
\[{{\text{t}}_{\text{r+1}}}\text{ = }\dfrac{{{\left( -1 \right)}^{\text{r}}}\text{n}\left( \text{n}-1 \right)\left( \text{n}-2 \right)\left( \text{n}-\text{r}+1 \right)}{\text{r}!}{{\text{x}}^{\text{r}}}\]
If \[\text{n}\]is any real number and \[\left| \text{b} \right| < \left| \text{a} \right|\], then
\[{{\text{(a+b)}}^{\text{n}}}={{\left[ \text{a}\left( 1+\dfrac{\text{b}}{\text{a}} \right) \right]}^{\text{n}}}\]
\[{{\text{(a+b)}}^{\text{n}}}={{\text{a}}^{\text{n}}}{{\left( 1+\dfrac{\text{b}}{\text{a}} \right)}^{\text{n}}}\]
Note:
While expanding \[{{\text{(a+b)}}^{\text{n}}}\] where \[\text{n}\]a negative integer or a fraction is, reduce the binomial to the form in which the first term is unity and the second term is numerically less than unity.
Particular expansion of the binomials for negative index, \[\left| \text{x} \right| < 1\]
$\dfrac{\text{1}}{\text{1+x}}\text{ = }{{\left( \text{1+x} \right)}^{\text{-1}}} \text{ = 1-x+}{{\text{x}}^{\text{2}}}\text{-}{{\text{x}}^{\text{3}}}\text{+}{{\text{x}}^{\text{4}}}\text{-}{{\text{x}}^{\text{5}}}\text{+}....$
$\dfrac{\text{1}}{\text{1-x}}\text{ = }{{\left( \text{1-x} \right)}^{\text{-1}}} \text{ = 1+x+}{{\text{x}}^{\text{2}}}\text{+}{{\text{x}}^{\text{3}}}\text{+}{{\text{x}}^{\text{4}}}\text{+}{{\text{x}}^{\text{5}}}\text{+}....$
$\dfrac{\text{1}}{{{\left( \text{1+x} \right)}^{2}}}\text{ = }{{\left( \text{1+x} \right)}^{\text{-2}}} \text{ = 1-2x+3}{{\text{x}}^{\text{2}}}\text{-4}{{\text{x}}^{\text{3}}}\text{+}.... $
$ \dfrac{\text{1}}{{{\left( \text{1-x} \right)}^{2}}}\text{ = }{{\left( \text{1-x} \right)}^{\text{-2}}} \text{ = 1+2x+3}{{\text{x}}^{\text{2}}}\text{+4}{{\text{x}}^{\text{3}}}\text{+}.... $
Binomial Coefficients:
The coefficients \[{}^{\text{n}}{{\text{C}}_{\text{0}}}\text{,}{}^{\text{n}}{{\text{C}}_{\text{1}}}\text{,}{}^{\text{n}}{{\text{C}}_{\text{2}}}\text{, }..\text{ ,}{}^{\text{n}}{{\text{C}}_{\text{n}}}\] in the expansion of \[{{\text{(a+b)}}^{\text{n}}}\] are called the binomial coefficients and denoted as \[{{\text{C}}_{0}},{{\text{C}}_{1}},{{\text{C}}_{2}},..,{{\text{C}}_{\text{n}}}\] respectively.
Now,
\[{{\left( \text{1+x} \right)}^{\text{n}}}\text{=}{}^{\text{n}}{{\text{C}}_{\text{0}}}{{\text{x}}^{\text{0}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{1}}}{{\text{x}}^{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}{{\text{x}}^{\text{2}}}\text{+}.....\text{+}{}^{\text{n}}{{\text{C}}_{\text{n}}}{{\text{x}}^{\text{n}}}.....\left( \text{i} \right)\]
Put \[\text{x}=1\]
\[{{\left( \text{1+1} \right)}^{\text{n}}}\text{=}{}^{\text{n}}{{\text{C}}_{\text{0}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}\text{+}.....\text{+}{}^{\text{n}}{{\text{C}}_{\text{n}}}\]
\[\therefore \text{ }{{\text{2}}^{\text{n}}}\text{ = }{}^{\text{n}}{{\text{C}}_{\text{0}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}\text{+}.....\text{+}{}^{\text{n}}{{\text{C}}_{\text{n}}}\]
\[\therefore {}^{\text{n}}{{\text{C}}_{\text{0}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}\text{+}.....\text{+}{}^{\text{n}}{{\text{C}}_{\text{n}}}={{\text{2}}^{\text{n}}}\]
\[\therefore {{\text{C}}_{\text{0}}}\text{+}{{\text{C}}_{\text{1}}}\text{+}{{\text{C}}_{\text{2}}}\text{+}.....\text{+}{{\text{C}}_{\text{n}}}={{\text{2}}^{\text{n}}}\]
Therefore, The sum of all binomial coefficients is \[{{\text{2}}^{\text{n}}}\]
Put \[\text{x}=-1\]in equation \[\left( \text{i} \right)\], we get
\[{{\left( \text{1-1} \right)}^{\text{n}}}\text{=}{}^{\text{n}}{{\text{C}}_{\text{0}}}-{}^{\text{n}}{{\text{C}}_{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}\text{- }.....\text{+}{{\left( -1 \right)}^{\text{n}}}{}^{\text{n}}{{\text{C}}_{\text{n}}}\]
\[\therefore \text{ 0 = }{}^{\text{n}}{{\text{C}}_{\text{0}}}-{}^{\text{n}}{{\text{C}}_{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}\text{- }.....\text{+}{{\left( -1 \right)}^{\text{n}}}{}^{\text{n}}{{\text{C}}_{\text{n}}}\]
\[\therefore {}^{\text{n}}{{\text{C}}_{\text{0}}}-{}^{\text{n}}{{\text{C}}_{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}\text{- }.....\text{+}{{\left( -1 \right)}^{\text{n}}}{}^{\text{n}}{{\text{C}}_{\text{n}}}=\text{0}\]
\[\therefore {}^{\text{n}}{{\text{C}}_{\text{0}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}\text{+}{}^{\text{n}}{{\text{C}}_{4}}\text{+ }.....={}^{\text{n}}{{\text{C}}_{\text{1}}}+{}^{\text{n}}{{\text{C}}_{3}}+{}^{\text{n}}{{\text{C}}_{5}}+.....\]
\[\therefore {{\text{C}}_{\text{0}}}\text{+}{{\text{C}}_{\text{2}}}\text{+}{{\text{C}}_{4}}\text{+ }.....={{\text{C}}_{\text{1}}}+{{\text{C}}_{3}}+{{\text{C}}_{5}}+.....\]
\[{{\text{C}}_{\text{0}}}\text{,}{{\text{C}}_{\text{2}}}\text{,}{{\text{C}}_{4}},...\] are known as even coefficients
\[{{\text{C}}_{1}}\text{,}{{\text{C}}_{3}}\text{,}{{\text{C}}_{5}},...\] are known as odd coefficients
Let,
\[{{\text{C}}_{\text{0}}}\text{+}{{\text{C}}_{\text{2}}}\text{+}{{\text{C}}_{4}}\text{+ }.....={{\text{C}}_{\text{1}}}+{{\text{C}}_{3}}+{{\text{C}}_{5}}+.....=\text{k}\]
Now,
\[{{\text{C}}_{\text{0}}}\text{+}{{\text{C}}_{\text{1}}}\text{+}{{\text{C}}_{\text{2}}}\text{+}.....\text{+}{{\text{C}}_{\text{n}}}={{\text{2}}^{\text{n}}}\]
Therefore,
\[\left( {{\text{C}}_{\text{0}}}\text{+}{{\text{C}}_{\text{2}}}\text{+}{{\text{C}}_{\text{4}}}\text{+ }..... \right)\text{ = }\left( {{\text{C}}_{\text{1}}}\text{+}{{\text{C}}_{\text{3}}}\text{+}{{\text{C}}_{\text{5}}}\text{+}..... \right)\text{ = }{{\text{2}}^{\text{n}}}\]
\[\text{k+k = }{{\text{2}}^{\text{n}}}\]
\[\text{2k = }{{\text{2}}^{\text{n}}}\]
\[\text{k = }\dfrac{{{\text{2}}^{\text{n}}}}{\text{2}}\]
\[\text{k = }{{\text{2}}^{\text{n-1}}}\]
\[\left( {{\text{C}}_{\text{0}}}\text{+}{{\text{C}}_{\text{2}}}\text{+}{{\text{C}}_{\text{4}}}\text{+ }..... \right)\text{ = }\left( {{\text{C}}_{\text{1}}}\text{+}{{\text{C}}_{\text{3}}}\text{+}{{\text{C}}_{\text{5}}}\text{+}..... \right)\text{ = }{{\text{2}}^{\text{n-1}}}\]
Therefore,
\[\text{The sum of even coefficients = The sum of odd coefficients = }{{\text{2}}^{\text{n-1}}}\]
Properties of Binomial Coefficient:
The coefficients have been omitted for the sake of simplicity.
\[{}^{\text{n}}{{\text{C}}_{\text{0}}}\text{,}{}^{\text{n}}{{\text{C}}_{\text{1}}}\text{,}{}^{\text{n}}{{\text{C}}_{\text{2}}}\text{, }..\text{ ,}{}^{\text{n}}{{\text{C}}_{\text{r}}},....,{}^{\text{n}}{{\text{C}}_{\text{n}}}\] are denoted by \[{{\text{C}}_{0}},{{\text{C}}_{1}},{{\text{C}}_{2}},..,{{\text{C}}_{\text{r}}}\text{,}....{{\text{C}}_{\text{n}}}\]
\[{{\text{C}}_{\text{0}}}\text{+}{{\text{C}}_{\text{1}}}\text{+}{{\text{C}}_{\text{2}}}\text{+}.....\text{+}{{\text{C}}_{\text{n}}}={{\text{2}}^{\text{n}}}\]
\[{{\text{C}}_{\text{0}}}\text{- }{{\text{C}}_{\text{1}}}\text{+}{{\text{C}}_{\text{2}}}\text{- }.....\text{+ }{{\left( -1 \right)}^{\text{n}}}{{\text{C}}_{\text{n}}}=\text{0}\]
\[{{\text{C}}_{\text{0}}}\text{+}{{\text{C}}_{\text{2}}}\text{+}{{\text{C}}_{4}}\text{+ }.....={{\text{C}}_{\text{1}}}+{{\text{C}}_{3}}+{{\text{C}}_{5}}+.....={{\text{2}}^{\text{n-1}}}\]
\[{}^{\text{n}}{{\text{C}}_{{{\text{r}}_{1}}}}={}^{\text{n}}{{\text{C}}_{{{\text{r}}_{2}}}}\Rightarrow {{\text{r}}_{\text{1}}}\text{=}{{\text{r}}_{\text{2}}}\text{ or }{{\text{r}}_{\text{1}}}\text{+}{{\text{r}}_{\text{2}}}\text{=n}\]
\[{}^{\text{n}}{{\text{C}}_{\text{r}}}+{}^{\text{n}}{{\text{C}}_{\text{r-1}}}={}^{\text{n+1}}{{\text{C}}_{\text{r}}}\]
\[\text{r}{}^{\text{n}}{{\text{C}}_{\text{r}}}=\text{n}{}^{\text{n-1}}{{\text{C}}_{\text{r}}}\]
Some Important Results:
\[{{\left( \text{1+x} \right)}^{\text{n}}}\text{=}{}^{\text{n}}{{\text{C}}_{\text{0}}}{{\text{x}}^{\text{0}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{1}}}{{\text{x}}^{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}{{\text{x}}^{\text{2}}}\text{+}.....\text{+}{}^{\text{n}}{{\text{C}}_{\text{n}}}{{\text{x}}^{\text{n}}}\]
Putting \[\text{x}=1\] and \[-1\], we get
\[{{\text{C}}_{\text{0}}}\text{+}{{\text{C}}_{\text{1}}}\text{+}{{\text{C}}_{\text{2}}}\text{+}.....\text{+}{{\text{C}}_{\text{n}}}={{\text{2}}^{\text{n}}}\]
\[{{\text{C}}_{\text{0}}}\text{- }{{\text{C}}_{\text{1}}}\text{+}{{\text{C}}_{\text{2}}}\text{- }.....\text{+ }{{\left( -1 \right)}^{\text{n}}}{{\text{C}}_{\text{n}}}=\text{0}\]
Differentiating \[{{\left( \text{1+x} \right)}^{\text{n}}}\text{=}{}^{\text{n}}{{\text{C}}_{\text{0}}}{{\text{x}}^{\text{0}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{1}}}{{\text{x}}^{\text{1}}}\text{+}{}^{\text{n}}{{\text{C}}_{\text{2}}}{{\text{x}}^{\text{2}}}\text{+}.....\text{+}{}^{\text{n}}{{\text{C}}_{\text{n}}}{{\text{x}}^{\text{n}}}\] , on both sides, \[\text{n}{{\left( \text{1+x} \right)}^{\text{n -1}}}\]
\[\text{= }{{\text{C}}_{1}}\text{+2}{{\text{C}}_{\text{1}}}\text{x+3}{{\text{C}}_{3}}{{\text{x}}^{2}}\text{+}.....\text{+n}{{\text{C}}_{\text{n}}}{{\text{x}}^{\text{n -1}}}.....\left( 1 \right)\]
\[\text{x=1}\]
\[\Rightarrow \text{n}{{\text{2}}^{\text{n-1}}}\text{ = }{{\text{C}}_{\text{1}}}\text{+2}{{\text{C}}_{\text{1}}}\text{+3}{{\text{C}}_{\text{3}}}\text{+}.....\text{+n}{{\text{C}}_{\text{n}}}\]
\[\text{x = -1}\]
\[\Rightarrow \text{0 = }{{\text{C}}_{\text{1}}}\text{- 2}{{\text{C}}_{\text{1}}}\text{+}.....\text{+}{{\left( \text{-1} \right)}^{\text{n -1}}}\text{n}{{\text{C}}_{\text{n}}}\]
Differentiating \[\left( 1 \right)\] again and again we will have different results.
Integrating \[{{\left( \text{1 + x} \right)}^{\text{n}}}\], we get
\[\dfrac{{{\left( \text{1+x} \right)}^{\text{n+1}}}}{\text{n+1}}\text{+C=}{{\text{C}}_{\text{0}}}\text{x+}\dfrac{{{\text{C}}_{\text{1}}}{{\text{x}}^{\text{2}}}}{\text{2}}\text{+}\dfrac{{{\text{C}}_{\text{2}}}{{\text{x}}^{\text{3}}}}{\text{3}}\text{+}....\text{+}\dfrac{{{\text{C}}_{\text{n}}}{{\text{x}}^{\text{n+1}}}}{\text{n+1}}\]
Where, \[\text{C}\] is a constant.
Put \[\text{x = 0}\], we get
\[\text{C = - }\dfrac{\text{1}}{\left( \text{n+1} \right)}\]
Therefore,
\[\dfrac{{{\left( \text{1+x} \right)}^{\text{n+1}}}\text{-1}}{\text{n+1}}\text{=}{{\text{C}}_{\text{0}}}\text{x+}\dfrac{{{\text{C}}_{\text{1}}}{{\text{x}}^{\text{2}}}}{\text{2}}\text{+}\dfrac{{{\text{C}}_{\text{2}}}{{\text{x}}^{\text{3}}}}{\text{3}}\text{+}....\text{+}\dfrac{{{\text{C}}_{\text{n}}}{{\text{x}}^{\text{n+1}}}}{\text{n+1}}.....\left( \text{2} \right)\]
Put \[\text{x = 1}\] in eqn. \[\left( 2 \right)\], we get
\[\dfrac{{{\text{2}}^{\text{n+1}}}\text{-1}}{\text{n+1}}\text{ = }{{\text{C}}_{\text{0}}}\text{+}\dfrac{{{\text{C}}_{\text{1}}}}{\text{2}}\text{+}\dfrac{{{\text{C}}_{\text{2}}}}{\text{3}}\text{+}....\text{+}\dfrac{{{\text{C}}_{\text{n}}}}{\text{n+1}}\]
Put \[\text{x = -1}\] in eqn.\[\left( 2 \right)\], we get
\[\dfrac{\text{1}}{\text{n+1}}\text{ = }{{\text{C}}_{\text{0}}}\text{- }\dfrac{{{\text{C}}_{\text{1}}}}{\text{2}}\text{+}\dfrac{{{\text{C}}_{\text{2}}}}{\text{3}}\text{ - }....\]
Example:
Find the coefficient of \[{{\text{x}}^{4}}\] in the expansion of \[\dfrac{\text{1+ x}}{1-\text{x}}\] if \[\left| \text{x} \right| < 1\]
Ans:
\[\dfrac{\text{1+ x}}{1-\text{x}}=\left( \text{1+ x} \right){{\left( \text{1- x} \right)}^{-1}}\]
\[=\left( \text{1+ x} \right)\left[ 1+\dfrac{\left( -1 \right)}{1!}\left( -\text{x} \right)+\dfrac{\left( -1 \right)\left( -1-1 \right)}{2!}{{\left( -\text{x} \right)}^{2}} +\dfrac{\left( -1 \right)\left( -1-1 \right)\left( -1-2 \right)}{3!}{{\left( -\text{x} \right)}^{3}}+...\infty \right]\]
\[=\left( \text{1+ x} \right)\left( 1+\text{x}+{{\text{x}}^{2}}+{{\text{x}}^{3}}+{{\text{x}}^{4}}+...\infty \right)\]
\[=\left( 1+\text{x}+{{\text{x}}^{2}}+{{\text{x}}^{3}}+{{\text{x}}^{4}}+...\infty \right)+\left( \text{x}+{{\text{x}}^{2}}+{{\text{x}}^{3}}+{{\text{x}}^{4}}+...\infty \right)\]
\[=1+2\text{x}+2{{\text{x}}^{2}}+2{{\text{x}}^{3}}+2{{\text{x}}^{4}}+...\infty \]
\[{{\text{x}}^{4}}=2\]
Hence, the coefficient of \[{{\text{x}}^{4}}\] is \[2\]
Multinomial Expansion:
In the expansion of \[{{\left( {{\text{x}}_{1}}+{{\text{x}}_{2}}+...+{{\text{x}}_{\text{n}}} \right)}^{\text{m}}}\] where \[\text{m,n }\in \text{ N}\] and \[{{\text{x}}_{1}}+{{\text{x}}_{2}}+...+{{\text{x}}_{\text{n}}}\] are independent variables.
We have
Total number of terms \[\text{=}{{\text{ }}^{\text{m+n -1}}}{{\text{C}}_{\text{n -1}}}\]
Coefficient of \[{{\text{x}}_{\text{1}}}^{^{{{\text{r}}_{\text{1}}}}}\text{ }{{\text{x}}_{\text{2}}}^{^{{{\text{r}}_{\text{2}}}}}\text{ }{{\text{x}}_{\text{3}}}^{{{\text{r}}_{\text{3}}}}....{{\text{x}}_{\text{n}}}^{{{\text{r}}_{\text{n}}}}\]
(Where, \[{{\text{r}}_{\text{1}}}+{{\text{r}}_{2}}+....+{{\text{r}}_{n}}=\text{m}\], \[{{\text{r}}_{\text{i}}}\in \text{N }\cup \left\{ 0 \right\}\]) is \[\dfrac{\text{m!}}{{{\text{r}}_{\text{1}}}\text{! }{{\text{r}}_{\text{2}}}\text{! }...\text{ }{{\text{r}}_{\text{n}}}\text{!}}\]
Sum of all the coefficients is obtained by putting all the variables \[{{\text{x}}_{\text{1}}}\] equal to \[1\]
Example:
Find the total number of terms in the expansion of ${{\left( \text{1+a+b} \right)}^{\text{10}}}$ and coefficient of ${{\text{a}}^{\text{2}}}{{\text{b}}^{\text{3}}}$.
Ans:
Total number of terms \[\text{=}{{\text{ }}^{\text{m+n -1}}}{{\text{C}}_{\text{n -1}}}\]
Therefore,
Total number of terms \[\text{=}{{\text{ }}^{\text{10+3 -1}}}{{\text{C}}_{\text{3 -1}}}\]
\[\text{=}{{\text{ }}^{\text{12}}}{{\text{C}}_{\text{2}}}\]
\[=66\]
Coefficient of ${{\text{a}}^{\text{2}}}{{\text{b}}^{\text{3}}}$can be calculated as,
Coefficient of ${{\text{a}}^{\text{2}}}{{\text{b}}^{\text{3}}}$\[=\dfrac{\text{10!}}{\text{2!}\times \text{3!}\times \text{5!}}\]
\[=2520\]
Class 11 Maths Revision Notes for Chapter-8 Binomial Theorem - Free PDF Download
Binomial Expression
The algebraic expression is called binomial, which includes only two terms. It is a polynomial with two terms. It is also known as a sum or difference between two or more monomials. It is the simplest polynomial form.
Binomial Theorem
Here are the necessary formula that you should always try and remember:
(x + y)ⁿ = ⁿCₒxⁿ⁻⁰ y⁰ + ⁿC₁xⁿ⁻¹y¹ + ⁿC₂xⁿ⁻²y² + ………
+ ⁿCᵣx \[^{n-r}\] y\[^{r}\] + ……. + ⁿC\[_{n-1}\] xy\[^{n-1}\] + ⁿC\[_{n}\]x⁰yⁿ
i.e.. (x + y)ⁿ = \[\sum_{r=0}^{n}\] ⁿCᵣ x\[^{n-1}\]y\[^{r}\] …..(i)
Here ⁿCₒ, ⁿC₁, ⁿC₂,...... ⁿC\[_{n}\] are called binomial coefficients and
ⁿCᵣ = \[\frac{n!}{r!(n-r)!}\] for 0 ≤ r ≤ n.
This is called the binomial theorem.
The various terms that you find in the nCx format are known are binomial coefficients.
Properties of the Binomial Expansion
Total number of terms in the expansion of (x + a)n is (n + 1).
The sum of the indices of x and a in each term is n.
It is a correct expansion when the terms are complex numbers.
Terms that are equidistant from both ends will have coefficients that are equal. These are termed differently - binomial co-efficients.
General term in the expansion of (x + c)n is given by Tr + 1 = nCrx n – r ar .
It is important to note that the values first increase and then decrease as you go ahead in the expansion.
The coefficient of xr in the expansion of (1+ x)n is nCr.
Middle Term in the Expansion of(1 + x)n
If n is even, then in the expansion of (x + a)n, the middle term is (n/2 + 1)th terms.
If n is odd, then in the expansion of (x + a)n , the middle terms are (n + 1) / 2 th term and (n + 3) / 2 th term.
Greatest Coefficient
If n is even, then in (x + a)n , the greatest coefficient is nCn / 2
If n is odd, then in (x + a)n , the greatest coefficient is nCn – 1 / 2 or nCn + 1 / 2 both being equal.
Fun Facts about the Greatest Term
Do you know that there are different ways to find the greatest term? You can do this by finding whether certain parts of the resulting expansion are integers or not, one can find the greatest term for each expansion. It is important to understand the definition of integers and remembering the greatest term’s formula for each of the cases.
Why Should You Download Revision Notes Class 11 Chapter 7 from Vedantu?
There are several reasons why you should download revision notes of class 11 chapter 7. A list of those reasons is mentioned below.
The revision notes will students be more prepared for their examination
Students will be able to score better grades and marks
Students can easily post their queries on the Vedantu platform and one of our experts will reply with the correct solution to the problem within 24 hours
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FAQs on Binomial Theorem Class 11 Notes CBSE Maths Chapter 7 (Free PDF Download)
1. Expand the Expression (1 – 2x)⁵?
From binomial theorem expansion, we can write as
(1 - 2x)⁵
= ⁵C₀(1)⁵ - ⁵C₁(1)⁴ (2x) + ⁵C₂(1)³(2x)² - ⁵C₃(1)²(2x)³ + ⁵C₄(1)¹(2x)⁴ - ⁵C₅(2x)⁵
= 1 - 5(2x) + 10(4x²) - 10(8x³) + 5(16x⁴) - (32x⁵)
= 1 - 10x + 40x² - 80x³ - 32x⁵
2. Which Number is Larger? The Two Numbers are (1.1) 10,000 and 1,000
Split the 1.1 and apply the binomial theorem, the first few terms of (1.1)¹⁰⁰⁰⁰, it is:
(1.1)¹⁰⁰⁰⁰ = (1 + 0.1)¹⁰⁰⁰⁰
= (1 + 0.1)10,000 C1(1.1) + other positive terms
= 1 + 10,000 x 1.1 + other positive terms
= 1 + 11,000 + other positive terms
>1000
(1.1)¹⁰⁰⁰⁰ > 1000
3. What is the procedure to download the PDF file of the revision notes of Chapter 7 of Class 11 Maths?
Beneath is the method for downloading the PDF file of the revision notes of Chapter 7 of Class 11 Maths:
Click on the given link CBSE Class 11 chapter 7.
After selecting the link, the official website of Vedantu will open.
Vedantu will provide students with the revision notes of Chapter 7 of Class 11 Maths.
Students will find that there is an option of “Download PDF” at the top of the Vedantu page.
Tap that option.
The revision notes will get downloaded.
4. What is meant by the term binomial expression?
According to Chapter 7 of Class 11 Maths, the expression in which two different terms are combined using operators like + or –is known as binomial expression.
For example, 100x – 50y, 200m + 100n
Now take an expression ( a + b )^n, in the expansion of this expression, the coefficient of the first term = the coefficient of the last term. Therefore, in this expansion, the first term and the last term are at equal distances and have the same coefficient.
5. State some properties of the Binomial Expression.
The properties of the Binomial Expression are stated below:
In the expansion of ( x + a )^n, the total number of terms is (n+1).
In each term, the sum of the indices of a and x is n.
For the terms of complex numbers, binomial expression is the correct expansion.
The coefficients are the same for the terms, which are equidistant from the ends. These coefficients are known as different binomial coefficients.
As you will move ahead in solving expansion, you will find that the values first increase and then decrease.
6. Give the overview of Chapter 7 of Class 11 Maths.
“Binomial Theorem” is Chapter 7 of Class 11 Maths. Through this chapter, students will learn about the Binomial Theorem for positive integers. Some calculations become difficult when students solve them using repeated multiplication. These complex calculations become easy when students will solve them using this binomial expression. The chapter also discusses the general and middle term, Pascal's Triangle in the binomial expansion and their applications. The chapter also consists of 2 exercises with miscellaneous exercises.
7. What is the best study plan for preparing Chapter 7 of Class 11 Maths?
From the examination point of view, Chapter 7 “Binomial Theorem” of Maths is a very important chapter for a student of Class 11. Students should learn and practice this chapter if they want to perform well in the Maths examination. For this, they can prefer the NCERT book as the concepts are explained in detail in this book. By solving the NCERT questions, students can comprehend the chapter easily. They can also use other reference books for clarifying their doubts.
If you want to understand these concepts more accurately, go through the given link CBSE Class 11 Chapter 7. This link will redirect you to the official website of Vedantu where you can access the content related to Chapter 7 for free. Additionally, you can also download its PDF if you want to study offline.
