已知数列{an}的通项公式为1/(n^2+4n+3),则其前n项和为多少?
问题描述：
已知数列{an}的通项公式为1/(n^2+4n+3),则其前n项和为多少?
答
{an}=1/(n^2+4n+3)=1/(n+1)(n+3)=1/2*(1/(n+1)-1/(n+3)),当n大于等于3时sn=a1+a2+……+an=1/2*(1/2-1/4+1/3-1/5+1/4-1/6+……+1/n-1/n+2+1/n+1-1/n+3）=1/2*（5/6-1/n+2—1/n+3）。当n=1或n=2时直接计算结果
答
{an}=1/(n+1)(n+3)=[1/(n+1)-1/(n+3)]/2
Sn=(1/2)[1/2-1/4+1/3-1/5+1/4-1/6+...+1/n-1/(n+2)+1/(n+1)-1/(n+3)]
=(1/2)[1/2+1/3-1/(n+2)-1/(n+3)]
=(1/2)[5/6-1/(n+2)-1/(n+3)] 其中n∈N