Question

急急急……已知数列An=n(n+2)，求数列｛1/An｝的前n项和Sn

已知数列An=n(n+2)，求数列｛1/An｝的前n项和Sn

Answer 1

1/An=1/n(n+2)=(n+2/n-n/n(n+2))/2=(1/n-1/(n+2))/2
所以Sn=(（1/1-1/3+1/2-1/4+1/3-1/5+1/4-1/6......+1/(n-2)-1/n+1/(n-1)-1/(n+1)+1/n-1/(n+2))/2=((1/1+1/2-1/(n+1)-1/(n+2)))/2=
不想通分了，就是这个形式吧。

Answer 2

Sn=1/2[(1-1/3)+(1/2-1/4)+(1/3-1/5)···+(1/(n-1)-1/(n+1))+(1/n-1/(n+2))]=1/2(3/2-1/(n+1)+1/(n)-1/(n+2))
最后再通分算一下即可。

Answer 3

解：1/An=1/(n(n+2))
=1/2(1/n-1/(n+2));
Sn=1/2((1-1/3)+(1/2-1/4)+……+（1/n-1/(n+2)）)=1/2(1+1/2-1/(n+1)-1/(n+2))
=1/2(3/2-1/(n+1)-1/(n+2));