First slide

Binomial theorem for positive integral Index

Question

if C_r stand forⁿC_r, the sum of the given series $\frac{2 (\frac{n}{2})! (\frac{n}{2})!}{n!} \cdot [C_{0}^{2} - 2 C_{1}^{2} + 3 C_{2}^{2} - \dots + (- 1)^{n} (n + 1) C_{n}^{2}]$ ,where n is an even positive integer, is

Moderate

A

0
B

$(- 1)^{n / 2} (n + 1)$
C

$(- 1)^{n} (n + 2)$
D

${(- 1)}^{\frac{n}{2}} (n + 2)$

Solution

We have, $C_{0}^{2} - 2 C_{1}^{2} + 3 C_{2}^{2} - \dots + (- 1)^{n} (n + 1) C_{n}^{2}$
$= [C_{0}^{2} - C_{1}^{2} + C_{2}^{2} - \dots + (- 1)^{n} C_{n}^{2}]$ $- [C_{1}^{2} - 2 C_{2}^{2} + 3 C_{3}^{2} - \dots + (- 1)^{n} {nC}_{n}^{2}]$
$= (- 1)^{n / 2} \frac{n!}{(\frac{n}{2})! (\frac{n}{2})!} \cdot - (- 1)^{n / 2 - 1} \cdot \frac{1}{2} n \frac{n!}{(\frac{n}{2})! (\frac{n}{2})!}$
$= (- 1)^{\frac{n}{2}} \cdot \frac{n!}{(\frac{n}{2})! (\frac{n}{2})!} \cdot (1 + \frac{n}{2})$
Therefore, the value of the given expression is $\frac{2 (\frac{n}{2})! (\frac{n}{2})!}{n!} \times (- 1)^{\frac{n}{2}} \cdot \frac{(n)!}{(\frac{n}{2})! (\frac{n}{2})!} (1 + \frac{n}{2})$
$= (- 1)^{\frac{n}{2}} (2 + n)$

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