Question

Let $S_n={\sum }_{r=0}^n{(-2)}^r\left(\frac{{{}^{n}C}_r}{{{}^{r+2}C}_r}\right)$, then

• A. $S_n=\frac{1}{n+1}$ if $n$ is odd
• B. $S_n=\frac{1}{n+2}$ if $n$ is odd
• C. $S_n=\frac{1}{n+1}$ if $n$ is even
• D. $S_n=\frac{1}{n+2}$ if $n$ is even

Answer 1

Answer: (B), (C)

$S_n=\frac{1}{n+2}$ if $n$ is odd
$S_n=\frac{1}{n+1}$ if $n$ is even

$\frac{{{}^{n}C}_r}{{{}^{r+2}C}_r}=\frac{!n}{r!(n-r)!}\frac{r!2!}{(r+2)!}=\frac{2(n!)}{(r+2)!(n-r)!}$
$\frac{2}{(n+1)(n+2)}\frac{(n+2)!}{(r+2)!\left[\right(n+2)-(r+2\left)\right]!}=\frac{2}{(n+1)(n+2)}={{}^{n+2}C}_{r+2}$
Thus,
$S={\sum }_{r=0}^n{(-2)}^r\left(\frac{{{}^{n}C}_r}{{{}^{r+2}C}_r}\right)=\frac{2}{(n+1)(n+2)}{\sum }_{r=0}^n{(-2)}^r{{}^{n+2}C}_{r+2}$
Putting $r+2=s$
$=\frac{2}{4(n+1)(n+2)}{\sum }_{s=2}^{n+2}{{}^{n+2}C}_s{(-2)}^s$
$=\frac{1}{2(n+1)(n+2)}\times [{\sum }_{s=0}^{n+2}{{}^{n+2}C}_s{(-2)}^s-{{}^{n+2}C}_0{(-2)}^0-{{}^{n+2}C}_1{(-2)}^1]$
$=\frac{1}{2(n+1)(n+2)}[{(1-2)}^{n+2}-1+2(n+2\left)\right]=\frac{1}{2(n+1)(n+2)}[2n+3+{(-1)}^n]$
But
$2n+3+{(-1)}^n=\left\{2\right(n+2)$ if $n$ is even

$2(n+1)$ if $n$ is odd$\}$
Thus
$S=\{\frac{1}{n+1}$, if $n$ is even

$\frac{1}{n+2}$, if $n$ is odd $\}$