已知数列an的前n项和为Sn=n^2+2n,求和:1/(a1*a2)+1/(a2*a3)+...+1/(an*a(n+1))_数学解答

已知数列an的前n项和为Sn=n^2+2n,求和:1/(a1*a2)+1/(a2*a3)+...+1/(an*a(n+1))

人气：142 ℃　时间：2020-01-29 22:55:00

解答

a1=S1=3,n>1时,an=Sn-S(n-1)=n^2+2n-(n-1)^2-2(n-1)=2n+1,a1=2*1+1=3,所以通项公式为an=2n-11/(a1a2)+1/(a2a3)+...+1/[ana(n+1)]=1/(3*5)+1/(5*7)+1/(7*9)+.+1/[(2n+1)(2n+3)]=1/3-1/5+1/5-1/7+1/7-1/9+.+1/(2n+1)-1...