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 [AIEEE 2004]

We have, \[{{S}_{N}}=\sum\limits_{r=0}^{n}{\frac{1}{^{n}{{C}_{r}}}}\] and \[{{t}_{n}}=\sum\limits_{r=0}^{n}{\frac{r}{^{n}{{C}_{r}}}}\] \[{{t}_{n}}=\sum\limits_{r=0}^{n}{\frac{n-(n-r)}{^{n}{{C}_{n-r}}}}\], \[[\because \,{{\,}^{n}}{{C}_{r}}={{\,}^{n}}{{C}_{n-r}}]\] = \[n\sum\limits_{r=0}^{n}{\frac{1}{^{n}{{C}_{r}}}}-\sum\limits_{r=0}^{n}{\frac{n-r}{^{n}{{C}_{n-r}}}}\] \[{{t}_{n}}=n\,.\,{{S}_{n}}-\left[ \frac{n}{^{n}{{C}_{n}}}+\frac{n-1}{^{n}{{C}_{n-1}}}+.....+\frac{1}{^{n}{{C}_{1}}}+0 \RIGHT]\] \[{{t}_{n}}=n\,.\,{{S}_{n}}-\sum\limits_{r=0}^{n}{\frac{r}{^{n}{{C}_{r}}}}\] \[\Rightarrow \] \[{{t}_{n}}=n\,.\,{{S}_{n}}-{{t}_{n}}\] \[\Rightarrow \] \[2{{t}_{n}}={{\,}^{n}}{{S}_{n}}\Rightarrow \frac{{{t}_{n}}}{{{S}_{n}}}=\frac{n}{2}\].