已知数列{an}的通项公式是an=1/(n^2+5n+6),则前n项和为
问题描述：
已知数列{an}的通项公式是an=1/(n^2+5n+6),则前n项和为
答
n^2+5n+6=(n+2)(n+3) an=1/(n+2)-1/(n+3)
an+a(n-1)=1/(n+2)-1/(n+3)+1/(n+1)-1(n+2)=1/(n+1)-1/(n+3)
Sn=a1+a2+a3......an=1/3-1/4+1/4-1/5+1/5-1/6......1/(n+1)-1/(n+2)+1/(n+2)-1/(n+3)
所以Sn =1/3-1/(n+3)
答
an=1/(n^2+5n+6),
an=1/(n+2)(n+3)=1/(n+2)-1/(n+3)
所以
Sn=a1+a2+a3+...+an
=1/3-1/4+1/4-1/5+...+1/(n+2)-1/(n+3)
=1/3-1/(n+3)
=n/3(n+3)