Question

If $n>2$ and $\alpha ,\beta ,\gamma \in R$ , then the value of $S=\alpha C_{0}-\left(\alpha + \beta \right)C_{1}$ $+\left(\alpha + 2 \beta + 2^{2} \gamma \right)C_{2}$ $-\left(\alpha + 3 \beta + 3^{2} \gamma \right)C_{3}+....$ upto $\left(n + 1\right)$ terms is equal to

(where, $C_{r}$ denotes $^{n}C_{r}$ )

• A. $0$
• B. $2^{n - 2}\gamma $
• C. $n^{2}2^{n - 2}\gamma $
• D. $n\gamma $

Answer 1

Answer: (D)

$S=\alpha\left(C_{0}-C_{1}+C_{2}-C_{3} \ldots \ldots\right)+\beta\left(-C_{1}+2 C_{2}-3 C_{3}+\ldots \ldots\right)$
$+\gamma\left(2^{2} C_{2}-3^{2} C_{3}+\ldots \ldots\right)$
$=\alpha \sum_{k=0}^{n}(-1)^{k}\left(C_{k}\right)+\beta \sum_{k=1}^{n}(-1)^{k}\left(k C_{k}\right)+\gamma \sum_{k=2}^{n}(-1)^{k}\left(k^{2} C_{k}\right)$
But, $\sum_{k=0}^{n}(-1)^{k} C_{k}=0$
$\sum_{k=1}^{n}(-1)^{k} k C_{k}=\left.\frac{d}{d x}(1-x)^{n}\right|_{x=1}=0$
and $\sum_{k=2}^{n}(-1)^{k} k^{2} C_{k}=\sum_{k=2}^{n}(-1)^{k}\{k(k-1)+k\} C_{k}$
$=\sum_{k=2}^{n}(-1)^{k} k(k-1) C_{k}+\sum_{k=1}^{n}(-1)^{k} k C_{k}+C_{1}$
$=0+0+n=n$
$\Rightarrow S=\alpha(0)+\beta(0)+\gamma(n)=n \gamma$