Question

The value of $\sum _{r=0}^n(-2{)}^r\left(\frac{{}^n{C}_r}{{}^{r+2}{C}_r}\right)$ is

• A. $\frac{1}{n+1},$ when $n$ is odd
• B. $\frac{1}{n+2},$ when $n$ is odd
• C. $\frac{1}{n+1},$ when $n$ is even
• D. $\frac{1}{n+2},$ when $n$ is even

Answer 1

Answer: (B), (C)

The correct options are
B $\frac{1}{n+2},$ when $n$ is odd
C $\frac{1}{n+1},$ when $n$ is even

For $r\ge 0,$ we have $\left(\frac{{}^n{C}_r}{{}^{r+2}{C}_r}\right)$
$=\frac{n!}{(n-r)!r!}\times \frac{r!2!}{(r+2)!}=\frac{n!\times 2!}{(n-r)!(r+2)!}$
$=\frac{2}{(n+1)(n+2)}\times \frac{(n+1)(n+2)n!}{\left\{\right(n+2)-(r+2\left)\right\}(r+2)!}$
$=\frac{2}{(n+1)(n+2)}\times {}^{n+2}{C}_{r+2}$

$\therefore \sum _{r=0}^n(-2{)}^r\left(\frac{{}^n{C}_r}{{}^{r+2}{C}_r}\right)$
$=\frac{2}{(n+1)(n+2)}\sum _{r=0}^n{}^{n+2}{C}_{r+2}(-2{)}^r$
$=\frac{1}{2(n+1)(n+2)}\sum _{r=0}^n{}^{n+2}{C}_{r+2}(-2{)}^{r+2}$

Putting $r+2=s$
$=\frac{1}{2(n+1)(n+2)}\sum _{s=2}^{n+2}{}^{n+2}{C}_s(-2{)}^s$
$=\frac{1}{2(n+1)(n+2)}\times \left[\sum _{s=0}^{n+2}{}^{n+2}{C}_s(-2{)}^s-{}^{n+2}{C}_0(-2{)}^0-{}^{n+2}{C}_1(-2{)}^1\right]$
$=\frac{1}{2(n+1)(n+2)}\left[\right(1-2{)}^{n+2}-1+2(n+2)]$
$=\{\begin{array}{cc}\frac{1}{2(n+1)(n+2)}(2n+4),& \text{if n is even}\\ \frac{1}{2(n+1)(n+2)}(2n+2),& \text{if n is odd}\end{array}$
$=\{\begin{array}{cc}\frac{1}{(n+1)},& \text{if n is even}\\ \frac{1}{(n+2)},& \text{if n is odd}\end{array}$