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Algebra Difficulty 2.8 Junior Find the answer

Given the sequence an=14n2−1(n∈N∗)a_{n}= \frac {1}{4n^{2}-1}(n\in\mathbb{N}^{*}), the sum of the first 1010 terms of the sequence {an}\{a_{n}\} is ()(\quad)

Pick one

Solution

Solution: The sequence an=14n2−1=12(12n−1−12n+1)a_{n}= \frac {1}{4n^{2}-1}= \frac {1}{2}( \frac {1}{2n-1}- \frac {1}{2n+1}),

∴Sn=12[(1−13)+(13−15)+…+(12n−1−12n+1)]=12(1−12n+1)=n2n+1\therefore S_{n}= \frac {1}{2}[(1- \frac {1}{3})+( \frac {1}{3}- \frac {1}{5})+\ldots+( \frac {1}{2n-1}- \frac {1}{2n+1})]= \frac {1}{2}(1- \frac {1}{2n+1})= \frac {n}{2n+1},

∴S10=1021\therefore S_{10}= \frac {10}{21}.

Therefore, the correct answer is: C\boxed{C}.

This can be derived using the "splitting terms for summation" method.
This question examines the "splitting terms for summation" method and is considered a basic question.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.