Question

已知数列的通项公式Bn=1/n(n+2),求数列的前n项和.

Answer 1

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

Sn=(1/2)[1/1-1/3+1/2-1/4+1/3-1/5+...+1/(n-1)-1/(n+1)+1/n-1(n+2)]

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

裂项求和.