数列1/（n²+3n+2）前n项和怎么算
问题描述：
数列1/（n²+3n+2）前n项和怎么算
答
an=1/(n+1)(n+2)
=[(n+2)-(n+1)]/(n+1)(n+2)
=(n+2)/(n+1)(n+2)-(n+1)/(n+1)(n+2)
=1/(n+1)-1/(n+2)
所以Sn=1/2-1/3+1/3-1/4+……+1/(n+1)-1/(n+2)
=1/2-1/(n+2)
=n/(2n+4)