An Alternating Reciprocal Sum

7/10RecursionalgebrasequencesTelescoping

Problem

A positive sequence {an}\{a_n\} satisfies a1=32a_1 = \dfrac{3}{2} and

an+12an2=1(n+2)21n2.a_{n+1}^2 - a_n^2 = \frac{1}{(n+2)^2} - \frac{1}{n^2}.

Let SnS_n be the sum of the first nn terms. Find

1S11S3+1S51S2007+1S2009.\frac{1}{S_1} - \frac{1}{S_3} + \frac{1}{S_5} - \cdots - \frac{1}{S_{2007}} + \frac{1}{S_{2009}}.

Answer

Solution

Difficulty7/10
TopicsRecursion, algebra, sequences, Telescoping

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