∑r=0n n−3r+1n−r+1 nCr2r is equal to

∑r=0n n−3r+1n−r+1 nCr2r=∑r=0n n−r+1-2rn−r+1 nCr2r; ∑r=0n nCr2r-2rn−r+1 nCr2r⇒=1+12n−rn−r+1× nCr2r−1∵ nCr nCr−1=n-r+1r=32n−∑r=1n nCr−12r−1=32n-1+12n-12n=12n

∑r=0n n−3r+1n−r+1 nCr2r is equal to

∑r=0n n−3r+1n−r+1 nCr2r=∑r=0n n−r+1-2rn−r+1 nCr2r; ∑r=0n nCr2r-2rn−r+1 nCr2r⇒=1+12n−rn−r+1× nCr2r−1∵ nCr nCr−1=n-r+1r=32n−∑r=1n nCr−12r−1=32n-1+12n-12n=12n

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$\sum_{r = 0}^{n} (\frac{n - 3 r + 1}{n - r + 1}) \frac{^{n} C_{r}}{2^{r}}$ is equal to

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

$\frac{1}{3^{n}}$

$\frac{1}{4^{n}}$

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

answer is A.

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Detailed Solution

$\begin{array}{l} \sum_{r = 0}^{n} \frac{n - 3 r + 1}{n - r + 1} \frac{^{n} C_{r}}{2^{r}} = \sum_{r = 0}^{n} (\frac{(n - r + 1) - 2 r}{n - r + 1}) \frac{^{n} C_{r}}{2^{r}}; \\ \sum_{r = 0}^{n} \frac{^{n} C_{r}}{2^{r}} - (\frac{2 r}{n - r + 1}) \frac{^{n} C_{r}}{2^{r}} \\ \Rightarrow= {(1 + \frac{1}{2})}^{n} - \frac{r}{n - r + 1} \times \frac{^{n} C_{r}}{2^{r - 1}} ∵ \frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{n - r + 1}{r} \\ = {(\frac{3}{2})}^{n} - \sum_{r = 1}^{n} \frac{^{n} C_{r - 1}}{2^{r - 1}} = {(\frac{3}{2})}^{n} - [{(1 + \frac{1}{2})}^{n} - \frac{1}{2^{n}}] = \frac{1}{2^{n}} \end{array}$