A) \[\frac{n}{2}\]
B) \[\frac{n}{2}-1\]
C) \[n-1\]
D) \[\frac{2n-1}{2}\]
Correct Answer: A
Solution :
Given that, \[{{s}_{n}}=\sum\limits_{r=0}^{n}{\frac{1}{^{n}{{C}_{r}}}}\] \[{{s}_{n}}=\sum\limits_{r=0}^{n}{\frac{1}{^{n}{{C}_{n-r}}}}\] \[({{\because }^{n}}{{C}_{r}}{{=}^{n}}{{C}_{n-r}})\] \[\Rightarrow \] \[n\,{{s}_{n}}=\sum\limits_{r=0}^{n}{\frac{n}{^{n}{{C}_{n-r}}}}\] \[\Rightarrow \] \[n\,{{s}_{n}}=\sum\limits_{r=0}^{n}{\left[ \frac{n-r}{^{n}{{C}_{n-r}}}+\frac{r}{^{n}{{C}_{n-r}}} \right]}\] \[\Rightarrow \] \[n\,{{s}_{n}}=\sum\limits_{r=0}^{n}{\frac{n-r}{^{n}{{C}_{n-r}}}}+\sum\limits_{r=0}^{n}{\frac{r}{^{n}{{C}_{r}}}}\] \[\Rightarrow \]\[n\,{{s}_{n}}=\left( \frac{n}{^{n}{{C}_{n}}}+\frac{n-1}{^{n}{{C}_{n-1}}}+....+\frac{1}{^{n}{{C}_{n}}} \right)+\sum\limits_{r=0}^{n}{\frac{r}{^{n}{{C}_{r}}}}\] \[\Rightarrow \] \[n{{s}_{n}}={{t}_{n}}+{{t}_{n}}\] \[\Rightarrow \] \[n{{s}_{n}}=2{{t}_{n}}\] \[\Rightarrow \] \[\frac{{{t}_{n}}}{{{s}_{n}}}=\frac{n}{2}\]
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