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Question:
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If $\mathrm{C}_{\mathrm{j}}$ stands for ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{j}}$, then
$\frac{\mathrm{C}_0}{2}+\frac{\mathrm{C}_1}{2.2^2}+\frac{\mathrm{C}_2}{3.2^3}+\ldots+\frac{\mathrm{C}_{\mathrm{n}}}{(\mathrm{n}+1) 2^{\mathrm{n}+1}}=$
Options:
$\frac{\mathrm{C}_0}{2}+\frac{\mathrm{C}_1}{2.2^2}+\frac{\mathrm{C}_2}{3.2^3}+\ldots+\frac{\mathrm{C}_{\mathrm{n}}}{(\mathrm{n}+1) 2^{\mathrm{n}+1}}=$
Solution:
2192 Upvotes
Verified Answer
The correct answer is:
$\frac{3^{n+1}}{2^{n+1}(n+1)}$
$\frac{C_0}{2}+\frac{C_1}{2.2^2}+\frac{C_2}{3.2^3}+\ldots+\frac{C_n}{(n+1) 2^{n+1}}$
$\begin{aligned} & =\sum_{r=0}^n \frac{{ }^n C_r}{r+1}\left(\frac{1}{2}\right)^{r+1} \\ & =\frac{1}{n+1} \sum_{r=0}^n \frac{n+1}{r+1}{ }^n \mathrm{C}_r\left(\frac{1}{2}\right)^{r+1}=\frac{1}{n+1} \sum_{r=0}^n{ }^{n+1} \mathrm{C}_{r+1}\left(\frac{1}{2}\right)^{r+1} \\ & \frac{1}{n+1}\left(1+\frac{1}{2}\right)^{n+1}=\frac{3^{n+1}}{(\mathrm{n}+1) 2^{\mathrm{n}+1}}\end{aligned}$
$\begin{aligned} & =\sum_{r=0}^n \frac{{ }^n C_r}{r+1}\left(\frac{1}{2}\right)^{r+1} \\ & =\frac{1}{n+1} \sum_{r=0}^n \frac{n+1}{r+1}{ }^n \mathrm{C}_r\left(\frac{1}{2}\right)^{r+1}=\frac{1}{n+1} \sum_{r=0}^n{ }^{n+1} \mathrm{C}_{r+1}\left(\frac{1}{2}\right)^{r+1} \\ & \frac{1}{n+1}\left(1+\frac{1}{2}\right)^{n+1}=\frac{3^{n+1}}{(\mathrm{n}+1) 2^{\mathrm{n}+1}}\end{aligned}$
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