数列的计算
问题描述：
数列的计算
数列1/1+2、1/1+2+3、1/1+2+3+4、……的前N项和为?
答
an=1/[1+2+……+(n+1)]=1/[(n+1)(n+2)/2]
=2/(n+1)(n+2)
=2*[1/(n+1)-1/(n+2)]
所以Sn=2*(1/2-1/3)+2*(1/3-1/4)+……+2*[1/(n+1)-1/(n+2)]
=2*[1/2-1/3+1/3-1/4+……+1/(n+1)-1/(n+2)]
=2*[1/2-1/(n+2)]
=1-2/(n+2)
=n/(n+2)