Question

If $\displaystyle\sum^n_{R = 0} (-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 \displaystyle\sum^{n}_{r=0} \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)} \displaystyle\sum^{n}_{r=0} \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$