If ${a_n} = {{ - 2} \over {4{n^2} - 16n + 15}}$, then ${a_1} + {a_2}\, + \,....\, + \,{a_{25}}$ is equal to :
Solution
<p>$$\sum\limits_{i = 1}^{25} {{a_i} = \sum {{{ - 2} \over {4{n^2} - 16n + 15}} = \sum {{{ - 2} \over {(2n - 5)(2n - 3)}}} } } $$</p>
<p>$$ = \sum\limits_{i = 1}^{25} {\left( {{1 \over {2n - 3}} - {1 \over {2n - 5}}} \right)} $$</p>
<p>$$ = \left[ {\left( {{1 \over { - 1}} - {1 \over { - 3}}} \right) + \left( {{1 \over 1} - {1 \over { - 1}}} \right) + \left( {{1 \over 3} - {1 \over 1}} \right)......} \right.$$</p>
<p>$= {1 \over {2(25) - 3}} + {1 \over 3} = {{50} \over {141}}$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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