If {a_n} = sumlimits_{r = 0}^n {} frac{1}{{^n | vedclass

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Question

(c) Given ${a_n} = \sum\limits_{r = 0}^n {} \frac{1}{{^n{C_r}}}$

Let ${b_n} = \sum\limits_{r = 0}^n {} \frac{r}{{^n{C_r}}}$

Then ${b_n} = \frac{

Vedclass · Accepted Answer

$\frac{1}{2}n{a_n}$

Vedclass · Answer

(c) Given ${a_n} = \sum\limits_{r = 0}^n {} \frac{1}{{^n{C_r}}}$

Let ${b_n} = \sum\limits_{r = 0}^n {} \frac{r}{{^n{C_r}}}$

Then ${b_n} = \frac{0}{{^n{C_0}}} + \frac{1}{{^n{C_1}}} + \frac{2}{{^n{C_2}}} + ........ + \frac{n}{{^n{C_n}}}$

and ${b_n} = \frac{n}{{^n{C_0}}} + \frac{{n - 1}}{{^n{C_1}}} + \frac{{n - 2}}{{^n{C_2}}} + .... + \frac{0}{{^n{C_n}}}$

By adding $2{b_n} = \frac{n}{{^n{C_0}}} + \frac{n}{{^n{C_1}}} + ...... + \frac{n}{{^n{C_n}}}$
$ = n\,\,\left[ {\frac{1}{{^n{C_0}}} + \frac{1}{{^n{C_1}}} + \frac{1}{{^n{C_2}}} + ...... + \frac{1}{{^n{C_n}}}} ight] \Rightarrow 2{b_n} = n{a_n}$

$	herefore $${b_n} = \frac{1}{2}n{a_n}$

If ${a_n} = \sum\limits_{r = 0}^n {} \frac{1}{{^n{C_r}}}$ then $\sum\limits_{r = 0}^n {} \frac{r}{{^n{C_r}}}$ equals

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