જો {S_n} = sumlimits_{r = 0}^n {frac{1}{{^n{C | vedclass

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Question

(d) We have, ${S_n} = \sum\limits_{r = 0}^n {\frac{1}{{^n{C_r}}}} $ and ${t_n} = \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} $

${t_n} = \sum\limits

Vedclass · Accepted Answer

$\frac{1}{2}n$

Vedclass · Answer

(d) We have, ${S_n} = \sum\limits_{r = 0}^n {\frac{1}{{^n{C_r}}}} $ and ${t_n} = \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} $

${t_n} = \sum\limits_{r = 0}^n {\frac{{n - (n - r)}}{{^n{C_{n - r}}}}} $,  $[\,{\,^n}{C_r} = {\,^n}{C_{n - r}}]$

= $n\sum\limits_{r = 0}^n {\frac{1}{{^n{C_r}}}} - \sum\limits_{r = 0}^n {\frac{{n - r}}{{^n{C_{n - r}}}}} $

${t_n} = n\,.\,{S_n} - \left[ {\frac{n}{{^n{C_n}}} + \frac{{n - 1}}{{^n{C_{n - 1}}}} + ..... + \frac{1}{{^n{C_1}}} + 0} \right]$

${t_n} = n\,.\,{S_n} - \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} $

$ \Rightarrow $ ${t_n} = n\,.\,{S_n} - {t_n}$ $ \Rightarrow $ $2{t_n} = {\,^n}{S_n} \Rightarrow \frac{{{t_n}}}{{{S_n}}} = \frac{n}{2}$.

જો ${S_n} = \sum\limits_{r = 0}^n {\frac{1}{{^n{C_r}}}} $ અને ${t_n} = \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} $, તો $\frac{{{t_n}}}{{{S_n}}}$ = . . .

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