求数列{1/(2n+1)(2n+3)}的前n项和参考书上解释到这一步我没看懂谁能解释一下这是怎么换算的an={1/(2n+1)(_数学解答

求数列{1/(2n+1)(2n+3)}的前n项和
参考书上解释到这一步我没看懂谁能解释一下这是怎么换算的
an={1/(2n+1)(2n+3)}=1/2{1/(2n+1)-1/(2n+3)}

人气：287 ℃　时间：2020-04-08 14:12:17

解答

an=1/[(2n+1)(2n+3)]
=[(2n+3)-(2n+1)]/[2(2n+1)(2n+3)]
=(2n+3)/[2(2n+1)(2n+3)]-(2n+1)/[2(2n+1)(2n+3)]
=1/[2(2n+1)]-1/[2(2n+3)]
=(1/2)[1/(2n+1)-1/(2n+3)]