If [a_n = - 2 4n^2 - 16n + 15], then [a_1 + a_2, + ,...., + ,a_25] is equal to :

Question

If $${a_n} = {{ - 2} \over {4{n^2} - 16n + 15}}$$, then $${a_1} + {a_2}\, + \,....\, + \,{a_{25}}$$ is equal to :

Accepted Answer

$$\sum\limits_{i = 1}^{25} {{a_i} = \sum {{{ - 2} \over {4{n^2} - 16n + 15}} = \sum {{{ - 2} \over {(2n - 5)(2n - 3)}}} } } $$

$$ = \sum\limits_{i = 1}^{25} {\left( {{1 \over {2n - 3}} - {1 \over {2n - 5}}} \right)} $$

$$ = \left[ {\left( {{1 \over { - 1}} - {1 \over { - 3}}} \right) + \left( {{1 \over 1} - {1 \over { - 1}}} \right) + \left( {{1 \over 3} - {1 \over 1}} \right)......} \right.$$

$$ = {1 \over {2(25) - 3}} + {1 \over 3} = {{50} \over {141}}$$

Solutions · 1