题目

2^2/(2^2-1) ,4^2/(4^2-1) ,6^2/(6^2-1) ,8^2/(8^2-1)……这个数列怎么求和?

提问时间：2020-08-09

答案

Sn=2^2/(2^2-1)+4^2/(4^2-1)+6^2/(6^2-1)+……+n^2/((2n)^2-1)
=[1+1/(2^2-1)]+[1+1/(4^2-1)]+[1+1/(6^2-1)]+……+[1+1/((2n)^2-1)]
=n+0.5×[2/(2-1)×(2+1)+2/(4-1)×(4+1)+2/(6-1)×(6+1)+……+2/(2n-1)×(2n+1)]
=n+0.5×[(3-1)/1×3+(5-3)/3×5+(7-5)/5×7+……+((2n+1)-(2n-1))/(2n-1)×(2n+1)]
=n+0.5×[(1/1-1/3)+(1/3-1/5)+(1/5-1/7)+……+(1/(2n-1)-1/(2n+1)]
=n+0.5×[1-1/(2n+1)]
=n+n/(2n+1)