Question

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) $0$

Answer 1

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:
$\begin{aligned} & 3!=3\times 2\times 1 \\\ & \Rightarrow 6 \\\ \end{aligned}$