frac{C_0}{1} + frac{C_1}{2} + frac{C_2}{3} + | vedclass

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(C) We know that $\frac{C_r}{r + 1} = \frac{1}{n + 1} \binom{n + 1}{r + 1}$.Substituting this into the given expression:$\sum_{r=0}^{n} \frac{C_r}{r +

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

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(C) We know that $\frac{C_r}{r + 1} = \frac{1}{n + 1} \binom{n + 1}{r + 1}$.Substituting this into the given expression:$\sum_{r=0}^{n} \frac{C_r}{r + 1} = \sum_{r=0}^{n} \frac{1}{n + 1} \binom{n + 1}{r + 1}$$= \frac{1}{n + 1} \sum_{r=0}^{n} \binom{n + 1}{r + 1}$Let $k = r + 1$. As $r$ goes from $0$ to $n$,$k$ goes from $1$ to $n + 1$.$= \frac{1}{n + 1} \sum_{k=1}^{n + 1} \binom{n + 1}{k}$Since $\sum_{k=0}^{m} \binom{m}{k} = 2^m$,we have $\sum_{k=1}^{m} \binom{m}{k} = 2^m - \binom{m}{0} = 2^m - 1$.Here $m = n + 1$,so the sum is $2^{n + 1} - 1$.Therefore,the expression equals $\frac{2^{n + 1} - 1}{n + 1}$.

$\frac{C_0}{1} + \frac{C_1}{2} + \frac{C_2}{3} + .... + \frac{C_n}{n + 1} = $

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