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If ${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}}}} $, then $\frac{{{t_n}}}{{{S_n}}}$ is equal to
$\frac{{2n - 1}}{2}$
$\frac{1}{2}n - 1$
$n - 1$
$\frac{1}{2}n$
Solution
(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 $ $2{t_n} = {\,^n}{S_n} \Rightarrow \frac{{{t_n}}}{{{S_n}}} = \frac{n}{2}$.
