Question

级数∑1/(n+1)(n+2)怎么求和.

Answer 1

1/(n+1)(n+2)

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

=(n+2)/(n+1)(n+2)-(n+1)/(n+1)(n+2)

=1/(n+1)-1/(n+2)

n是从1开始到n吗?

原式=1/2-1/3+1/3-1/4+……+1/(n+1)-1/(n+2)

=1/2-1/(n+2)

=n/(2n+4)

你以为书上答案都是对的?