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यदि ${a_n} = \sum\limits_{r = 0}^n {} \frac{1}{{^n{C_r}}}$ है, तो $\sum\limits_{r = 0}^n {} \frac{r}{{^n{C_r}}}$ =
$(n - 1)\;{a_n}$
$n{a_n}$
$\frac{1}{2}n{a_n}$
इनमें से कोई नहीं
Solution
(c) दिया है, ${a_n} = \sum\limits_{r = 0}^n {} \frac{1}{{^n{C_r}}}$
माना कि ${b_n} = \sum\limits_{r = 0}^n {} \frac{r}{{^n{C_r}}}$
तो ${b_n} = \frac{0}{{^n{C_0}}} + \frac{1}{{^n{C_1}}} + \frac{2}{{^n{C_2}}} + …….. + \frac{n}{{^n{C_n}}}$
और ${b_n} = \frac{n}{{^n{C_0}}} + \frac{{n – 1}}{{^n{C_1}}} + \frac{{n – 2}}{{^n{C_2}}} + …. + \frac{0}{{^n{C_n}}}$
जोड़ने पर, $2{b_n} = \frac{n}{{^n{C_0}}} + \frac{n}{{^n{C_1}}} + …… + \frac{n}{{^n{C_n}}}$
$ = 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}$
