Question

求2^2+1/2^2-1,3^2+1/3^2-1,...(n+1)^2+1/(n+1)^2-1 的前n项的和Sn

Answer 1

(n+1)^2+1/(n+1)^2-1
=1+2/[(n+1)^2-1]
=1+2/[n(n+2)]
=1+1/n-1/(n+2)
故Sn=n+(1-1/3)+(1/2-1/4)+(1/3-1/5)+....+[1/n-1/(n+2)]
=n+1+1/2-1/(n+1)-1/(n+2)
=n+3/2-(2n+3)/(n+1)(n+2)
希望能帮到你，祝学习进步O(∩_∩)O

Answer 2

这个可以分组求和

(n+1)^2
借助公式：

1^2+2^2+…+n^2=n*(n+1)*(2n+1)/6

但1/(n+1)^2的前n项和，除了n趋于正无穷，否则没有确切的表达式