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Q.

If ${}^{n}C_{0}, {}^{n}C_{1}, {}^{n}C_{2}, - - - - {}^{n}C_{n}$ are the coefficients of the expansion of ${(1 + x)}^{n}$ then the value of $\sum_{0}^{k} \frac{{}^{n}C_{k}}{k + 1}$ is_________

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a

$\frac{2^{n} - 1}{n}$

b

$\frac{2^{n + 1} - 1}{n + 1}$

c

0

d

$\frac{1}{2}$

answer is C.

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Detailed Solution

$T_{r + 1} = \frac{^{n} C_{r}}{r + 1} But^{n + 1} C_{r + 1} = \frac{n + 1}{r + 1} \cdot^{n} C_{r} \Rightarrow T_{r + 1} = \frac{1}{n + 1} \cdot^{n + 1} C_{r + 1}$
$\sum_{0}^{n} \frac{^{n} C_{k}}{k + 1} = \frac{1}{n + 1} [^{n + 1} C_{1} +^{n + 1} C_{2} + - - - - +^{n + 1} C_{n + 1}]$
$= \frac{1}{n + 1} [2^{n + 1} -^{n + 1} C_{0}] = \frac{2^{n + 1} - 1}{n + 1}$

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