JEE MAIN - Mathematics Hindi (2022 - 28th June Evening Shift - No. 24)
माना $$\mathrm{n}=1,2, \ldots .50$$ के लिए अनंत गुणोत्तर श्रेढ़ी जिसका पहला पद $$\mathrm{n}^{2}$$ तथा सार्व अनुपात $$\frac{1}{(\mathrm{n}+1)^{2}}$$ है का योग $$S_{n}$$ है। तो $${1 \over {26}} + \sum\limits_{n = 1}^{50} {\left( {{S_n} + {2 \over {n + 1}} - n - 1} \right)} $$ का मान बराबर है ___________.
Answer
41651
Explanation
$${S_n} = {{{n^2}} \over {1 - {1 \over {{{(n + 1)}^2}}}}} = {{n{{(n + 1)}^2}} \over {n + 2}} = ({n^2} + 1) - {2 \over {n + 2}}$$
अब $${1 \over {26}} + \sum\limits_{n = 1}^{50} {\left( {{S_n} + {2 \over {n + 1}} - n - 1} \right)} $$
$$ = {1 \over {26}} + \sum\limits_{n = 1}^{50} {\left\{ {({n^2} - n) + 2\left( {{1 \over {n + 1}} - {1 \over {n + 2}}} \right)} \right\}} $$
$$ = {1 \over {26}} + {{50 \times 51 \times 101} \over 6} - {{50 \times 51} \over 2} + 2\left( {{1 \over 2} - {1 \over {52}}} \right)$$
$$ = 1 + 25 \times 17(101 - 3)$$
$$ = 41651$$
Comments (0)
