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