Question

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

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

Answer 1

Answer: (B)

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