frac{{^nC_0}}{1} + frac{{^nC_2}}{3} + frac{{^ | vedclass

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(C) Let $S = \sum_{k=0, 2, 4, \dots} \frac{{^nC_k}}{k+1}$.Using the identity $\frac{{^nC_k}}{k+1} = \frac{1}{n+1} {^{n+1}C_{k+1}}$,we have:$S = \frac{

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

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(C) Let $S = \sum_{k=0, 2, 4, \dots} \frac{{^nC_k}}{k+1}$.Using the identity $\frac{{^nC_k}}{k+1} = \frac{1}{n+1} {^{n+1}C_{k+1}}$,we have:$S = \frac{1}{n+1} \sum_{k=0, 2, 4, \dots} {^{n+1}C_{k+1}}$.Let $N = n+1$. Then $S = \frac{1}{N} \sum_{j=1, 3, 5, \dots} {^NC_j}$.The sum of odd-indexed binomial coefficients is $2^{N-1}$.Thus,$S = \frac{1}{N} \cdot 2^{N-1} = \frac{{2^n}}{n+1}$.Verification: For $n=2$,$S = \frac{{^2C_0}}{1} + \frac{{^2C_2}}{3} = 1 + \frac{1}{3} = \frac{4}{3}$.Option $(c)$ gives $\frac{{2^2}}{2+1} = \frac{4}{3}$,which matches.

$\frac{{^nC_0}}{1} + \frac{{^nC_2}}{3} + \frac{{^nC_4}}{5} + \frac{{^nC_6}}{7} + \dots = $

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