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What is the sum of the series:

$$\sum_{n=4}^\infty \frac{1}{n^2-2n-3}$$

(Hint: Use partial fractions to rewrite $\frac{1}{n^2-2n-3}$ as the difference of two fractions, and then show that the series is telescoping.)

A

$\cfrac{25}{48}$

B

$\cfrac{27}{55}$

C

$\cfrac{5}{4}$

D

$\cfrac{25}{12}$

E

$\cfrac{5}{2}$

Accuracy 0%
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