If {C_r} stands for ^n{C_r},the sum of the se | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $S = \sum_{r=0}^{n} (-1)^r (r+1) C_r^2$. We know that $\sum_{r=0}^{n} (-1)^r C_r^2 = (-1)^{n/2} \frac{n!}{(n/2)!(n/2)!}$ for even $n$. Also,$\

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $S = \sum_{r=0}^{n} (-1)^r (r+1) C_r^2$. We know that $\sum_{r=0}^{n} (-1)^r C_r^2 = (-1)^{n/2} \frac{n!}{(n/2)!(n/2)!}$ for even $n$. Also,$\sum_{r=0}^{n} (-1)^r r C_r^2 = (-1)^{n/2} \frac{n}{2} \frac{n!}{(n/2)!(n/2)!}$. Thus,$S = \sum_{r=0}^{n} (-1)^r C_r^2 + \sum_{r=0}^{n} (-1)^r r C_r^2$. $S = (-1)^{n/2} \frac{n!}{(n/2)!(n/2)!} + (-1)^{n/2} \frac{n}{2} \frac{n!}{(n/2)!(n/2)!}$. $S = (-1)^{n/2} \frac{n!}{(n/2)!(n/2)!} (1 + n/2) = (-1)^{n/2} \frac{n!}{(n/2)!(n/2)!} \frac{n+2}{2}$. Multiplying by the coefficient $\frac{2(n/2)!(n/2)!}{n!}$,we get: $\frac{2(n/2)!(n/2)!}{n!} 	imes (-1)^{n/2} \frac{n!}{(n/2)!(n/2)!} \frac{n+2}{2} = (-1)^{n/2}(n+2)$.

If ${C_r}$ stands for $^n{C_r}$,the sum of the series $\frac{{2(n/2)!(n/2)!}}{{n!}}[C_0^2 - 2C_1^2 + 3C_2^2 - ..... + {( - 1)^n}(n + 1)C_n^2]$,where $n$ is an even positive integer,is

Similar Questions

$\frac{{^nC_0}}{1} + \frac{{^nC_2}}{3} + \frac{{^nC_4}}{5} + \frac{{^nC_6}}{7} + \dots = $

Let $(1+2x)^{20} = a_0 + a_1x + a_2x^2 + \dots + a_{20}x^{20}$. Then $3a_0 + 2a_1 + 3a_2 + 2a_3 + 3a_4 + 2a_5 + \dots + 2a_{19} + 3a_{20}$ equals

If $(1 - x + x^2)^n = a_0 + a_1x + a_2x^2 + .... + a_{2n}x^{2n}$,then $a_0 + a_2 + a_4 + .... + a_{2n} = $

The sum $\sum\limits_{i = 0}^m {\binom{10}{i}} {\binom{20}{m - i}}$,(where $\binom{p}{q} = 0$ if $p < q$),is maximum when $m$ is

In the expansion of $(1 + x)^{50}$,the sum of the coefficients of odd powers of $x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module