A) \[(n-1){{a}_{n}}\]
B) \[n{{a}_{n}}\]
C) \[\frac{1}{2}n{{a}_{n}}\]
D) none of these
Correct Answer: C
Solution :
Given\[,\] \[{{a}_{n}}=\sum\limits_{r=0}^{1}{\frac{1}{^{n}{{C}_{r}}}}\] ? (i) Let \[{{b}_{n}}=\sum\limits_{r=0}^{b}{\frac{r}{^{n}{{C}_{r}}}}\] \[\Rightarrow \] \[{{b}_{n}}=\frac{0}{^{n}{{C}_{0}}}+\frac{1}{^{n}{{C}_{1}}}+\frac{1}{^{n}{{C}_{2}}}+...\] \[+\frac{n-1}{^{n}{{C}_{n-1}}}+\frac{n}{^{n}{{C}_{n}}}\] ? (ii) Or it can be rewritten as \[{{b}_{n}}=\frac{n}{^{n}{{C}_{n}}}+\frac{n-1}{^{n}{{C}_{n-1}}}+...+\frac{1}{^{n}{{C}_{n}}}+\frac{0}{^{n}{{C}_{0}}}\] ... (iii) On adding Eqs. (ii) and (iii), we get \[2{{b}_{n}}=\frac{n}{^{n}{{C}_{0}}}+\frac{n}{^{n}{{C}_{1}}}+...+\frac{n}{^{n}{{C}_{n}}}\] \[=n\left[ \frac{1}{^{n}{{C}_{0}}}+\frac{1}{^{n}{{C}_{1}}}+...+\frac{1}{^{n}{{C}_{n}}} \right]\] \[=n{{a}_{n}}\] [from (i)] \[\Rightarrow \] \[{{b}_{n}}=\frac{1}{2}n{{a}_{n}}\]
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