If a n = - 2 4 n 2 - 16 n + 15 , then a 1 + a 2 + … . + a 25 is equal to:

We have,

$a_n=-\frac{2}{4n^2-16n+15}$

$\Rightarrow a_n=-\frac{2}{\left(2n-5\right)\left(2n-3\right)}$

$\Rightarrow a_n=-\frac{1}{2}\left[\frac{1}{\left(n-{\displaystyle \frac{5}{2}}\right)\left(n-\frac{3}{2}\right)}\right]$

$\Rightarrow a_n=-\frac{1}{2}\left[\frac{1}{\left(n-{\displaystyle \frac{5}{2}}\right)}-\frac{1}{\left(n-{\displaystyle \frac{3}{2}}\right)}\right]$

$\Rightarrow a_n=\frac{1}{2}\left[\frac{1}{\left(n-{\displaystyle \frac{3}{2}}\right)}-\frac{1}{\left(n-{\displaystyle \frac{5}{2}}\right)}\right]$

$\Rightarrow \sum _{n=1}^{25}{a}_n=\frac{1}{2}\sum _{n=1}^{25}\left[\frac{1}{\left(n-{\displaystyle \frac{3}{2}}\right)}-\frac{1}{\left(n-{\displaystyle \frac{5}{2}}\right)}\right]$

$\Rightarrow \sum _{n=1}^{25}{a}_n=\frac{1}{2}\left[\left\{\frac{1}{\left(-{\displaystyle \frac{1}{2}}\right)}-\frac{1}{\left(-{\displaystyle \frac{3}{2}}\right)}\right\}+\left\{\frac{1}{\left({\displaystyle \frac{1}{2}}\right)}-\frac{1}{\left(-{\displaystyle \frac{1}{2}}\right)}\right\}+\left\{\frac{1}{\left({\displaystyle \frac{3}{2}}\right)}-\frac{1}{\left({\displaystyle \frac{1}{2}}\right)}\right\}+\left\{\frac{1}{\left({\displaystyle \frac{5}{2}}\right)}-\frac{1}{\left({\displaystyle \frac{3}{2}}\right)}\right\}+\left\{\frac{1}{\left({\displaystyle \frac{7}{2}}\right)}-\frac{1}{\left({\displaystyle \frac{5}{2}}\right)}\right\}+...+\left\{\frac{1}{\left({\displaystyle \frac{45}{2}}\right)}-\frac{1}{\left({\displaystyle \frac{43}{2}}\right)}\right\}+\left\{\frac{1}{\left({\displaystyle \frac{47}{2}}\right)}-\frac{1}{\left({\displaystyle \frac{45}{2}}\right)}\right\}\right]$

$\Rightarrow \sum _{n=1}^{25}{a}_n=\frac{1}{2}\left(\frac{2}{3}+\frac{2}{47}\right)$

$\Rightarrow \sum _{n=1}^{25}{a}_n=\left(\frac{1}{3}+\frac{1}{47}\right)$

$\Rightarrow \sum _{n=1}^{25}{a}_n=\frac{50}{141}$