Question

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

• A. $\frac{52}{147}$
• B. $\frac{50}{141}$
• C. $\frac{51}{144}$
• D. $\frac{49}{138}$

Answer 1

Answer: (B)

If $a_n=\frac{-2}{4 n^2-16 n+15}$ then $a_1+a_2+\ldots \ldots \ldots a_{25}$
$\Rightarrow \displaystyle\sum_{n=1}^{25} a _{ n }=\sum \frac{-2}{4 n ^2-16 n +15} $
$=\sum \frac{-2}{4 n^2-6 n-10 n +15} $
$=\sum \frac{-2}{2 n (2 n -3)-5(2 n -3)} $
$=\sum \frac{-2}{(2 n-3)(2 n-5)}$
$ =\sum \frac{1}{2 n-3}-\frac{1}{2 n-5} $
$=\frac{1}{47}-\frac{1}{(-3)} $
$ =\frac{50}{141}$