题：设数列{An}的通项公式为An=1/n2+4n+3,则其前n项的和为多少?5/12-1/2n+4-1/2n+6_数学解答

题：设数列{An}的通项公式为An=1/n2+4n+3,则其前n项的和为多少?
5/12-1/2n+4-1/2n+6

人气：284 ℃　时间：2020-03-28 15:20:11

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)