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Q. 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

JamiaJamia 2007

Solution:

Given that, $ {{s}_{n}}=\sum\limits_{r=0}^{n}{\frac{1}{^{n}{{C}_{r}}}} $ $ {{s}_{n}}=\sum\limits_{r=0}^{n}{\frac{1}{^{n}{{C}_{n-r}}}} $ $ ({{\because }^{n}}{{C}_{r}}{{=}^{n}}{{C}_{n-r}}) $ $ \Rightarrow $ $ n\,{{s}_{n}}=\sum\limits_{r=0}^{n}{\frac{n}{^{n}{{C}_{n-r}}}} $ $ \Rightarrow $ $ n\,{{s}_{n}}=\sum\limits_{r=0}^{n}{\left[ \frac{n-r}{^{n}{{C}_{n-r}}}+\frac{r}{^{n}{{C}_{n-r}}} \right]} $ $ \Rightarrow $ $ n\,{{s}_{n}}=\sum\limits_{r=0}^{n}{\frac{n-r}{^{n}{{C}_{n-r}}}}+\sum\limits_{r=0}^{n}{\frac{r}{^{n}{{C}_{r}}}} $ $ \Rightarrow $ $ n\,{{s}_{n}}=\left( \frac{n}{^{n}{{C}_{n}}}+\frac{n-1}{^{n}{{C}_{n-1}}}+....+\frac{1}{^{n}{{C}_{n}}} \right)+\sum\limits_{r=0}^{n}{\frac{r}{^{n}{{C}_{r}}}} $ $ \Rightarrow $ $ n{{s}_{n}}={{t}_{n}}+{{t}_{n}} $ $ \Rightarrow $ $ n{{s}_{n}}=2{{t}_{n}} $ $ \Rightarrow $ $ \frac{{{t}_{n}}}{{{s}_{n}}}=\frac{n}{2} $