数列通项公式为an=1/n2+4n+3,则其前n项的和为?_数学解答

数列通项公式为an=1/n2+4n+3,则其前n项的和为?

人气：299 ℃　时间：2020-02-01 01:16:51

解答

an=1/(n+1)(n+3)=1/2*[1/(n+1)-1/(n+3)]
所以Sn=1/2*[1/2-1/4+1/3-1/5+……+1/n-1/(n+2)+1/(n+1)-1/(n+3)]
=1/2*[1/2+1/3-1/(n+2)-1/(n+3)]
=(5n^2+13n)/(2n^2+10n+12)