- Home
- Standard 11
- Mathematics
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
$(n - 1)\;{a_n}$
$n{a_n}$
$\frac{1}{2}n{a_n}$
None of these
Solution
(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}}}} \right] \Rightarrow 2{b_n} = n{a_n}$
$\therefore $${b_n} = \frac{1}{2}n{a_n}$
