The sum of the series 1 + frac{1}{2} {}^{n}C_ | vedclass

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Question

(A) Let the sum be $S = \sum_{k=0}^{n} \frac{1}{k+1} {}^{n}C_{k}$.Using the identity $\frac{1}{k+1} {}^{n}C_{k} = \frac{1}{n+1} {}^{n+1}C_{k+1}$, we h

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

Vedclass · Answer

(A) Let the sum be $S = \sum_{k=0}^{n} \frac{1}{k+1} {}^{n}C_{k}$.Using the identity $\frac{1}{k+1} {}^{n}C_{k} = \frac{1}{n+1} {}^{n+1}C_{k+1}$, we have:$S = \sum_{k=0}^{n} \frac{1}{n+1} {}^{n+1}C_{k+1}$$S = \frac{1}{n+1} \sum_{k=0}^{n} {}^{n+1}C_{k+1}$$S = \frac{1}{n+1} [{}^{n+1}C_{1} + {}^{n+1}C_{2} + \dots + {}^{n+1}C_{n+1}]$Since $\sum_{r=0}^{m} {}^{m}C_{r} = 2^{m}$, we know that $\sum_{r=1}^{m} {}^{m}C_{r} = 2^{m} - {}^{m}C_{0} = 2^{m} - 1$.Here $m = n+1$, so the sum is $2^{n+1} - 1$.Therefore, $S = \frac{2^{n+1}-1}{n+1}$.

The sum of the series $1 + \frac{1}{2} {}^{n}C_{1} + \frac{1}{3} {}^{n}C_{2} + \dots + \frac{1}{n+1} {}^{n}C_{n}$ is equal to

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