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

Solution :
$$\eqalign{ & {S_n} = \frac{1}{{^n{C_0}}} + \frac{1}{{^n{C_1}}} + \frac{1}{{^n{C_2}}} + ..... + \frac{1}{{^n{C_n}}} \cr & {t_n} = \frac{0}{{^n{C_0}}} + \frac{1}{{^n{C_1}}} + \frac{2}{{^n{C_2}}} + ..... + \frac{n}{{^n{C_n}}} \cr & {t_n} = \frac{n}{{^n{C_n}}} + \frac{{n - 1}}{{^n{C_{n - 1}}}} + \frac{{n - 2}}{{^{^n}{C_{n - 2}}}} + ..... + \frac{0}{{^n{C_0}}} \cr & {\text{Add,}}\,\,{{2}}{{{t}}_n} = \left( n \right)\left[ {\frac{1}{{^n{C_0}}} + \frac{1}{{^n{C_1}}} + .....\frac{1}{{^n{C_n}}}} \right] = n{S_n} \cr & \therefore \,\,\frac{{{t_n}}}{{{S_n}}} = \frac{n}{2} \cr} $$