题目

已知数列是等差数列,且a1=3,a1+a2+a3=15,求数列的通项公式
（2）求{1/An*An+1}的前n项和Sn

提问时间：2021-01-02

答案

1.
设an=3+(n-1)d
15=a1+a2+a3
=3+3+d+3+2d
=9+3d
d=2
an=3+2(n-1)=2n+1;
2.
1/[ana(n+1)]=1/[(2n+1)(2n+3)]
=(1/2)[1/(2n+1)-1/(2n+3)]
即
2/[ana(n+1)]=1/(2n+1)-1/(2n+3)
2/[a(n-1)an]=1/(2n-1)-1/(2n+1)
2/[a(n-2)a(n-1)]=1/(2n-3)-1/(2n-1)
……
2/[a3a4]=1/7-1/9
2/[a2a3]=1/5-1/7
2/[a1a2]=1/3-1/5
两边相加：
2Sn=2/[ana(n+1)]+2/[a(n-1)an]+2/[a(n-2)a(n-1)]+……+2/[a3a4]+2/[a2a3]+2/[a1a2]
=1/3-1/(2n+3)
=[(2n+3)-3]/[3(2n+3)]=2n/[3(2n+3)]
2Sn=2n/[3(2n+3)]
所以
Sn=n/[3(2n+3)].