If Sn=∑r=0n1 nCr,tn=∑r=0nr nCr then tnSn=

tn=0 nC0+1 nC1+2 nC2+….+n nCn….(1)tn=n nC0+n−1 nC1+….+0 nCn….(2)(1)+(2)⇒2tn=nSn∴tnSn=n2

If Sn=∑r=0n1 nCr,tn=∑r=0nr nCr then tnSn=

tn=0 nC0+1 nC1+2 nC2+….+n nCn….(1)tn=n nC0+n−1 nC1+….+0 nCn….(2)(1)+(2)⇒2tn=nSn∴tnSn=n2

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If $S_{n} = \sum_{r = 0}^{n} \frac{1}{^{n} C_{r}}, t_{n} = \sum_{r = 0}^{n} \frac{r}{^{n} C_{r}}$ then $\frac{t_{n}}{S_{n}} =$

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$\frac{n}{2}$

answer is B.

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Detailed Solution

$\begin{array}{l} t_{n} = \frac{0}{^{n} C_{0}} + \frac{1}{^{n} C_{1}} + \frac{2}{^{n} C_{2}} + \dots . + \frac{n}{^{n} C_{n}} \dots . (1) \\ t_{n} = \frac{n}{^{n} C_{0}} + \frac{n - 1}{^{n} C_{1}} + \dots . + \frac{0}{^{n} C_{n}} \dots . (2) \\ (1) + (2) \Rightarrow 2 t_{n} = n (S_{n}) \\ ∴ \frac{t_{n}}{S_{n}} = \frac{n}{2} \end{array}$