Question

Prove that if n and r are positive integers
${n^r} - n{(n - 1)^r} + \dfrac{{n(n - 1)}}{{2!}}{(n - 2)^r} - \dfrac{{n(n - 1)(n - 2)}}{{3!}}{(n - 3)^r} + \cdots$
is equal to 0 if r be less than n, and to n! if r=n.

Answer 1

Hint-To solve this question, we need to know the basics of Chapter Binomial Theorem. First you should remember the binomial expansion which is given as ${(x + y)^n} = \sum\limits_{r = 0}^n {{x^{n - r}}} {y^r}$, where, n ∈ N and x,y,∈ R then
${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n}}!}}{{{\text{[r}}!{\text{(n - r)}}!{\text{]}}}}$
This is also called the binomial theorem formula which is used for solving many problems related to the expressions having a large number of terms.

Complete step-by-step solution:
Given in the question,
We have two positive integers n and r.
Given expression is:
${n^r} - n{(n - 1)^r} + \dfrac{{n(n - 1)}}{{2!}}{(n - 2)^r} - \dfrac{{n(n - 1)(n - 2)}}{{3!}}{(n - 3)^r} + \cdots$
Now, as we know,
The Binomial Theorem is the method of expanding an expression which has been raised to any finite power.
Therefore, we use the expansion of ${e^x}$.
${e^x}$= $\left( {1 + x + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + \cdots } \right)$
By manipulating this we get,
${({e^x} - 1)^n} = {\left( {x + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + \cdots } \right)^n}$
= ${x^n} +$ terms containing higher powers of x ………..... (1)
Again, by the Binomial Theorem,
${({e^x} - 1)^n} = {e^n}x - n{e^{(n - 1)x}} + \dfrac{{n(n - 1)}}{{1.2}}{e^{(n - 2)x}} - \cdots$ ……………. (2)
By expanding each of the terms ${e^{nx}}$, ${e^{(n - 1)x}}$, …. we find that the coefficient of ${x^r}$in above expression of ${({e^x} - 1)^n}$.
$\dfrac{1}{{r!}}$(${n^r} - n{(n - 1)^r} + \dfrac{{n(n - 1)}}{{2!}}{(n - 2)^r} - \dfrac{{n(n - 1)(n - 2)}}{{3!}}{(n - 3)^r} + \cdots$)
and by equating the coefficients of ${x^r}$in (1) and (2) the result follows.
The value of the given expression is equal to 0 if r be less than n, and to n! if r=n.
Thus, overall, we verify and satisfy the given condition of the question.

Note: The total number of terms in the expansion of ${\left( {x + y} \right)^n}$ are (n+1). The sum of exponents of x and y is always n. and The binomial coefficients which are equidistant from the beginning and from the ending are equal i.e.${}^{\text{n}}{{\text{C}}_0}$=${}^{\text{n}}{{\text{C}}_{\text{n}}}$, ${}^{\text{n}}{{\text{C}}_1}$ = ${}^{\text{n}}{{\text{C}}_{{\text{n - 1}}}}$ ….. etc.