A) \[-\frac{1}{n}\]
B) \[\frac{1}{n+1}\]
C) \[\frac{-1}{n+1}\]
D) \[\frac{n}{n+1}\]
Correct Answer: D
Solution :
[d] \[\frac{1}{n+1}\sum\limits_{r=1}^{n}{{{(-1)}^{r+1}}.\frac{n+1}{r+1}{{.}^{n}}{{C}_{r}}}\] \[=\frac{1}{n+1}\sum\limits_{r=1}^{n}{{{(-1)}^{r+1}}{{.}^{n+1}}{{C}_{r+1}}}\] \[=\frac{1}{n+1}{{[}^{n+1}}{{C}_{2}}{{-}^{n+1}}{{C}_{3}}{{+}^{n+1}}{{C}_{4}}-....+{{(-1)}^{n+1}}{{.}^{n+1}}{{C}_{n+1}}]\] \[=\frac{1}{n+1}{{[}^{n+1}}{{C}_{0}}{{-}^{n+1}}{{C}_{1}}{{+}^{n+1}}{{C}_{2}}-....+{{(-1)}^{n+1}}{{.}^{n+1}}{{C}_{n+1}}+n]\]\[=\frac{1}{n+1}[0+n]=\frac{n}{n+1}\]
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