Question

The sum of $\frac{nC_{r}}{r + 2C_{r}}$ is -

• A. $\frac{2}{n + 1}$
• B. $\frac{1}{n + 2}$
• C. $\frac{2}{n - 2}$
• D. $\frac{2}{n + 2}$

Answer 1

Answer: (D)

$\frac{2}{n + 2}$

Answer 2

We have, $\sum_{r = 0}^{n}{(–1)^{r}}$ $\frac{nC_{r}}{r + 2C_{r}}$

= $\sum _ { \mathrm { r } = 0 } ^ { \mathrm { n } } ( - 1 ) ^ { \mathrm { r } }$ $\frac{n!}{(n - r)!r!}$× $\frac{2!r!}{(r + 2)!}$

= 2 $\sum _ { \mathrm { r } = 0 } ^ { \mathrm { n } } ( - 1 ) ^ { \mathrm { r } }$ $\frac{n!}{(n - r)!(r + 2)!}$

= $\frac { 2 } { ( \mathrm { n } + 1 ) ( \mathrm { n } + 2 ) }$ $\sum_{r = 0}^{n}{(–1)^{r}}$

$\frac{(n + 2)!}{\{(n + 2)–(r + 2)\}!(r + 2)!}$

= $\frac { 2 } { ( \mathrm { n } + 1 ) ( \mathrm { n } + 2 ) }$ $\sum_{r = 0}^{n}{(–1)^{r + 2}}n + 2C_{r + 2}$

= $\frac { 2 } { ( \mathrm { n } + 1 ) ( \mathrm { n } + 2 ) }$ $\sum_{s = 2}^{n + 2}{(–1{)^{s}}^{n + 2}C_{s}}$

= $\frac { 2 } { ( \mathrm { n } + 1 ) ( \mathrm { n } + 2 ) }$ $\left\lbrack \left( \sum_{s = 0}^{n + 2}{(–1{)^{s}}^{n + 2}C_{s}} \right) - \left( n + 2C_{0}–^{n + 2}C_{1} \right) \right\rbrack$

= $\frac{2}{n + 2}$.

Hence (4) is correct answer.