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Question: \(\sum_{r = 0}^{n}{( - 1)^{r}\frac{r}{r + 1}}.^{n}C_{r}\) is equal to...

∑r=0n(−1)rrr+1.nCr\sum_{r = 0}^{n}{( - 1)^{r}\frac{r}{r + 1}}.^{n}C_{r} is equal to

A

1n+1\frac{1}{n + 1}

B

1n\frac{1}{n}

C

−1n+1- \frac{1}{n + 1}

D

−1n- \frac{1}{n}

Answer

−1n+1- \frac{1}{n + 1}

Explanation

Solution

∑r=0n(−1)rrr+1.nCr\sum_{r = 0}^{n}{( - 1)^{r}\frac{r}{r + 1}}.^{n}C_{r} = ∑r=0n(−1)rr+1−1r+1.nCr\sum_{r = 0}^{n}{( - 1)^{r}\frac{r + 1 - 1}{r + 1}}.^{n}C_{r}

= ∑r=0n(−1)rnCr−∑r=0n1r+1⋅nCr\sum _ { r = 0 } ^ { n } ( - 1 ) ^ { r } { } ^ { n } C _ { r } - \sum _ { r = 0 } ^ { n } \frac { 1 } { r + 1 } \cdot { } ^ { n } C _ { r }

= 0−∑r=0n(−1)rn+1.n+1r+1.nCr0 - \sum_{r = 0}^{n}{\frac{( - 1)^{r}}{n + 1}.\frac{n + 1}{r + 1}.^{n}C_{r}}

= −∑r=0n(−1)rn+1.n+1Cr+1- \sum_{r = 0}^{n}{\frac{( - 1)^{r}}{n + 1}.^{n + 1}C_{r + 1}}

= −1n+1∑r=0n(−1)rn+1Cr+1- \frac{1}{n + 1}\sum_{r = 0}^{n}{( - 1{)^{r}}^{n + 1}C_{r + 1}}

= −1n+1.n+1C0=−1n+1- \frac{1}{n + 1}.^{n + 1}C_{0} = - \frac{1}{n + 1}