Let for n = 1, 2, ......, 50, Sn be the sum of the infinite geometric progression whose first term is n2 and whose common ratio is ${1 \over {{{(n + 1)}^2}}}$. Then the value of
$${1 \over {26}} + \sum\limits_{n = 1}^{50} {\left( {{S_n} + {2 \over {n + 1}} - n - 1} \right)} $$ is equal to ___________.
Answer (integer)
41651
Solution
<p>$${S_n} = {{{n^2}} \over {1 - {1 \over {{{(n + 1)}^2}}}}} = {{n{{(n + 1)}^2}} \over {n + 2}} = ({n^2} + 1) - {2 \over {n + 2}}$$</p>
<p>Now $${1 \over {26}} + \sum\limits_{n = 1}^{50} {\left( {{S_n} + {2 \over {n + 1}} - n - 1} \right)} $$</p>
<p>$$ = {1 \over {26}} + \sum\limits_{n = 1}^{50} {\left\{ {({n^2} - n) + 2\left( {{1 \over {n + 1}} - {1 \over {n + 2}}} \right)} \right\}} $$</p>
<p>$$ = {1 \over {26}} + {{50 \times 51 \times 101} \over 6} - {{50 \times 51} \over 2} + 2\left( {{1 \over 2} - {1 \over {52}}} \right)$$</p>
<p>$= 1 + 25 \times 17(101 - 3)$</p>
<p>$= 41651$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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