Question

If $\sum _{R=0}^n(-1{)}^r\frac{{}^n{C}_r}{{}^{r+3}{C}_r}=\frac{3}{a+3}$ , then a - n is equal to

• A. 0
• B. 1
• C. 2
• D. None of these

Answer 1

Answer: (A)

$\frac{{}^n{C}_r}{{}^{r+3}{C}_r}=3!\frac{1}{\left(r+3\right)\left(r+2\right)}.\frac{{}^n{C}_r}{\left(r+1\right)}$
$=3!\frac{1}{\left(r+3\right)\left(r+2\right)}.\frac{{}^{n+1}{C}_{r+1}}{\left(n+1\right)}$
$=3!\frac{1}{\left(r+3\right)\left(n+1\right)}\frac{{}^{n+1}{C}_{r+1}}{r+2}$
$=3!\frac{1}{\left(r+3\right)\left(n+1\right)}\frac{{}^{n+2}{C}_{r+2}}{n+2}$
$=\frac{3!}{\left(n+1\right)\left(n+2\right)}.\frac{{}^{n+2}{C}_{r+2}}{r+3}$
$=\frac{3!}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}{}^{n+3}{C}_{r+3}$
$\therefore \sum _{r=0}^n{\left(-1\right)}^r\frac{{}^n{C}_r}{{}^{r+3}{C}_r}$
$=\frac{6}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\sum _{r=0}^n{\left(-1\right)}^r{}^{n+3}{C}_{r+3}$
$=\frac{6}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\left[{}^{n+3}{C}_3-{}^{n+3}{C}_4+...+{\left(-1\right)}^n{}^{n+3}{C}_{n+3}\right]$
$=\frac{6}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\left[{}^{n+3}{C}_0-{}^{n+3}{C}_1+{}^{n+3}{C}_2\right]$
$\left[\because {}^{n+3}{C}_0-{}^{n+3}{C}_1+......+{\left(-1\right)}^{n+3}\times {}^{n+3}{C}_{n+3}=0\right]$
$=\frac{6}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\left(1-n-3+\frac{\left(n+3\right)\left(n+2\right)}{2}\right)$
$=\frac{3}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\left(n^2+3n+2\right)$
$=\frac{3}{n+3}\frac{3}{n+3}$
$=\frac{3}{a+3}\Rightarrow n$
$=a\Rightarrow a-n$
$=0$