Answer
Verified
439.8k+ views
Hint: We will solve this problem by using binomial theorem. For this, let us understand what a binomial theorem given by the formula${{\left( x+a \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}}{{x}^{n-r}}{{a}^{r}}$. Then, by taking $ab$ common from the entire equation and by using the concept of combination to replace values like 1, n and so on as ${}^{n}{{C}_{0}}=1,{}^{n}{{C}_{1}}=n,{}^{n}{{C}_{2}}=\dfrac{n\left( n-1 \right)}{2}$ and${}^{n}{{C}_{n}}=1$, we get the desired form. Then, by cancelling the common terms from numerator and denominator, we get the required answer.
Complete step by step answer:
We will solve this problem by using binomial theorem. For this, let us understand what a binomial theorem is.
According to Binomial theorem, if $x$ and $a$are real numbers then for all\[n\in N\],
${{\left( x+a \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{a}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{a}^{1}}+{}^{n}{{C}_{2}}{{x}^{n-2}}{{a}^{2}}+...+{}^{n}{{C}_{r}}{{x}^{n-r}}{{a}^{r}}+...+{}^{n}{{C}_{n-1}}{{x}^{1}}{{a}^{n-1}}+{}^{n}{{C}_{n}}{{x}^{0}}{{a}^{n}}$ , i.e.,
${{\left( x+a \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}}{{x}^{n-r}}{{a}^{r}}$.
Now let us solve the given problem based on the Binomial theorem.
We have
$ab-n\left( a-1 \right)\left( b-1 \right)+\dfrac{n\left( n-1 \right)\left( a-2 \right)\left( b-2 \right)...+{{\left( -1 \right)}^{n}}\left( a-n \right)\left( b-n \right)}{1.2}$
Then, by taking $ab$ common from the entire equation, we get:
\[ab\left\{ 1-n\left( 1-\dfrac{1}{a} \right)\left( 1-\dfrac{1}{b} \right)+\dfrac{n\left( n-1 \right)}{1.2}\left( 1-\dfrac{2}{a} \right)\left( 1-\dfrac{2}{b} \right)...+{{\left( -1 \right)}^{n}}\left( 1-\dfrac{n}{a} \right)\left( 1-\dfrac{n}{b} \right) \right\}\]
Then, by using the concept of combination to replace values like 1, n and so on as:
${}^{n}{{C}_{0}}=1,{}^{n}{{C}_{1}}=n,{}^{n}{{C}_{2}}=\dfrac{n\left( n-1 \right)}{2}$ and${}^{n}{{C}_{n}}=1$.
Then, by substituting the above values in the expression, we get:
$ab\left\{ {}^{n}{{C}_{0}}1-{}^{n}{{C}_{1}}\left( 1-\dfrac{1}{a} \right)\left( 1-\dfrac{1}{b} \right)+{}^{n}{{C}_{2}}\left( 1-\dfrac{2}{a} \right)\left( 1-\dfrac{2}{b} \right)...+{{\left( -1 \right)}^{n}}{}^{n}{{C}_{n}}\left( 1-\dfrac{n}{a} \right)\left( 1-\dfrac{n}{b} \right) \right\}$
$\Rightarrow ab\left\{ \dfrac{{{a}^{n}}-{{b}^{n}}}{ab\left( a-b \right)} \right\}$
Now, by cancelling the common terms from numerator and denominator, we get:
$\dfrac{{{a}^{n}}-{{b}^{n}}}{\left( a-b \right)}$
So, we get the value of the given expression as $\dfrac{{{a}^{n}}-{{b}^{n}}}{\left( a-b \right)}$.
So, the correct answer is “Option B”.
Note: Now, to solve these type of problems we need to know the basics of the combination given by the formula as for ${}^{n}{{C}_{r}}$ as $^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$. Now, we must also know the formula for calculating the factorial of any number n by multiplying with (n-1) till it reaches 1. To understand let us find factorial of 3 as:
$\begin{align}
& 3!=3\times 2\times 1 \\
& \Rightarrow 6 \\
\end{align}$
Complete step by step answer:
We will solve this problem by using binomial theorem. For this, let us understand what a binomial theorem is.
According to Binomial theorem, if $x$ and $a$are real numbers then for all\[n\in N\],
${{\left( x+a \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{a}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{a}^{1}}+{}^{n}{{C}_{2}}{{x}^{n-2}}{{a}^{2}}+...+{}^{n}{{C}_{r}}{{x}^{n-r}}{{a}^{r}}+...+{}^{n}{{C}_{n-1}}{{x}^{1}}{{a}^{n-1}}+{}^{n}{{C}_{n}}{{x}^{0}}{{a}^{n}}$ , i.e.,
${{\left( x+a \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}}{{x}^{n-r}}{{a}^{r}}$.
Now let us solve the given problem based on the Binomial theorem.
We have
$ab-n\left( a-1 \right)\left( b-1 \right)+\dfrac{n\left( n-1 \right)\left( a-2 \right)\left( b-2 \right)...+{{\left( -1 \right)}^{n}}\left( a-n \right)\left( b-n \right)}{1.2}$
Then, by taking $ab$ common from the entire equation, we get:
\[ab\left\{ 1-n\left( 1-\dfrac{1}{a} \right)\left( 1-\dfrac{1}{b} \right)+\dfrac{n\left( n-1 \right)}{1.2}\left( 1-\dfrac{2}{a} \right)\left( 1-\dfrac{2}{b} \right)...+{{\left( -1 \right)}^{n}}\left( 1-\dfrac{n}{a} \right)\left( 1-\dfrac{n}{b} \right) \right\}\]
Then, by using the concept of combination to replace values like 1, n and so on as:
${}^{n}{{C}_{0}}=1,{}^{n}{{C}_{1}}=n,{}^{n}{{C}_{2}}=\dfrac{n\left( n-1 \right)}{2}$ and${}^{n}{{C}_{n}}=1$.
Then, by substituting the above values in the expression, we get:
$ab\left\{ {}^{n}{{C}_{0}}1-{}^{n}{{C}_{1}}\left( 1-\dfrac{1}{a} \right)\left( 1-\dfrac{1}{b} \right)+{}^{n}{{C}_{2}}\left( 1-\dfrac{2}{a} \right)\left( 1-\dfrac{2}{b} \right)...+{{\left( -1 \right)}^{n}}{}^{n}{{C}_{n}}\left( 1-\dfrac{n}{a} \right)\left( 1-\dfrac{n}{b} \right) \right\}$
$\Rightarrow ab\left\{ \dfrac{{{a}^{n}}-{{b}^{n}}}{ab\left( a-b \right)} \right\}$
Now, by cancelling the common terms from numerator and denominator, we get:
$\dfrac{{{a}^{n}}-{{b}^{n}}}{\left( a-b \right)}$
So, we get the value of the given expression as $\dfrac{{{a}^{n}}-{{b}^{n}}}{\left( a-b \right)}$.
So, the correct answer is “Option B”.
Note: Now, to solve these type of problems we need to know the basics of the combination given by the formula as for ${}^{n}{{C}_{r}}$ as $^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$. Now, we must also know the formula for calculating the factorial of any number n by multiplying with (n-1) till it reaches 1. To understand let us find factorial of 3 as:
$\begin{align}
& 3!=3\times 2\times 1 \\
& \Rightarrow 6 \\
\end{align}$
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE