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# If $n > 3$ and $a,b\in R$, then the value of $ab-n\left( a-1 \right)\left( b-1 \right)+\dfrac{n\left( n-1 \right)\left( a-2 \right)\left( b-2 \right)...+{{\left( -1 \right)}^{n}}\left( a-n \right)\left( b-n \right)}{1.2}$ is equal to (a) ${{a}^{n}}+{{b}^{n}}$ (b) $\dfrac{{{a}^{n}}-{{b}^{n}}}{a-b}$ (c) ${{\left( ab \right)}^{n}}$ (d) $0$

Last updated date: 13th Jun 2024
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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.

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}