If a_n = -24n^2-16n+15, then a_1+a_2++a_25 is equal to:

Question

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

Accepted Answer

$4n^2-16n+15 = (2n-3)(2n-5)$, so
$$a_n = \frac{-2}{(2n-3)(2n-5)} = \frac{1}{2n-3}-\frac{1}{2n-5}.$$

This telescopes:
$$\sum_{n=1}^{25}a_n = \frac{1}{47} - \frac{1}{-3} = \frac{1}{47}+\frac13 = \frac{50}{141}.$$

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