Question

求代数题解题思路

如图，这道题怎么解？谢谢！

Answer 1

本题解法思路是拆项后相互抵消法。
通项公式 ak = 1/[(2k-1)(2k+1)] = (1/2)[1/(2k-1) - 1/(2k+1)]， k = 1，2，... , 2017.

原式 = (1/2)[1-1/3 + 1/3-1/5 + 1/5-1/7 + ... + 1/2013-1/2015 + 1/2015-1/2017]
= (1/2)[1-1/2017] = (1/2)2016/2017 = 1008/2017

追问

谢谢，我知道啦。其实就是找出它们之间的规律。

Answer 2

an
=1/[(2n-1)(2n+1)]
=(1/2)[ 1/(2n-1) -1/(2n+1)]
Sn
=a1+a2+...+an
=(1/2)[ 1 -1/(2n+1)]
=n/(2n+1)
1/(1x3)+1/(3x5)+...+1/(2015x2017)
=S1008
=1008/(2(1008)+1)
=1008/2017

追问

亲，没看懂嘛意思……，不过还是要谢谢你