MCQ
If $a_n=\frac{-2}{4 n^2-16 n+15}$, then $a_1+a_2+\ldots \ldots+a_{25}$ is equal to :
- A$\frac{51}{144}$
- B$\frac{49}{138}$
- ✓$\frac{50}{141}$
- D$\frac{52}{147}$
✓
Answer
Correct option: C.
$\frac{50}{141}$
c
If $a_n=\frac{-2}{4 n^2-16 n+15}$ then $a_1+a_2+\ldots \ldots \ldots a_{25}$
If $a_n=\frac{-2}{4 n^2-16 n+15}$ then $a_1+a_2+\ldots \ldots \ldots a_{25}$
$\sum \limits_{n=1}^{25} a_n=\sum \frac{-2}{4 n^2-16 n+15}$
$=\sum \frac{-2}{4 n^2-6 n-10 n+15}$
$=\sum \frac{-2}{2 n(2 n-3)-5(2 n-3)}$
$=\sum \frac{-2}{(2 n-3)(2 n-5)}$
$=\sum \frac{1}{2 n-3}-\frac{1}{2 n-5}$
$=\frac{1}{47}-\frac{1}{(-3)}$
$=\frac{50}{141}$
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