If C_{j} stands for { }^{n} C_{j},then frac{C | vedclass

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(B) The given expression is $S = \sum_{r=0}^n \frac{{ }^n C_r}{r+1} \left(\frac{1}{2}\right)^{r+1}$.Using the property $\frac{{ }^n C_r}{r+1} = \frac{

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$\frac{3^{n+1} - 1}{2^{n+1}(n+1)}$

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(B) The given expression is $S = \sum_{r=0}^n \frac{{ }^n C_r}{r+1} \left(\frac{1}{2}\right)^{r+1}$.Using the property $\frac{{ }^n C_r}{r+1} = \frac{{ }^{n+1} C_{r+1}}{n+1}$,we get:$S = \frac{1}{n+1} \sum_{r=0}^n { }^{n+1} C_{r+1} \left(\frac{1}{2}\right)^{r+1}$.Let $k = r+1$,then $S = \frac{1}{n+1} \sum_{k=1}^{n+1} { }^{n+1} C_k \left(\frac{1}{2}\right)^k$.Using the binomial expansion $(1+x)^m = \sum_{k=0}^m { }^m C_k x^k$,we have $\sum_{k=1}^{n+1} { }^{n+1} C_k \left(\frac{1}{2}\right)^k = (1 + \frac{1}{2})^{n+1} - { }^{n+1} C_0 (\frac{1}{2})^0 = (\frac{3}{2})^{n+1} - 1$.Thus,$S = \frac{1}{n+1} [(\frac{3}{2})^{n+1} - 1] = \frac{3^{n+1} - 2^{n+1}}{(n+1) 2^{n+1}}$.

If $C_{j}$ stands for ${ }^{n} C_{j}$,then $\frac{C_0}{2} + \frac{C_1}{2 \cdot 2^2} + \frac{C_2}{3 \cdot 2^3} + \ldots + \frac{C_{n}}{(n+1) 2^{n+1}} = $

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