How do I find the sum below?


I think there should be a simpler way instead of just adding the ten terms up using brute force, since it's impossible to do that in less than 10 minutes in a contest, because the terms starting on the 4th one is ugly, and even uglier to get the common denominator.

Can someone give me some idea for this?

  • $\begingroup$ Just notice that you have $(i+1)^2-i^2$ there. $\endgroup$ Sep 19, 2013 at 13:35

1 Answer 1


Note that $$ \frac{2i+1}{i^{2} (i+1)^{2}} = \frac{1}{i^{2}} - \frac{1}{(i+1)^{2}}. $$

See under telescoping series.


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