Question

已知:a1+a2+a3...+an=n^3;

已知数列an满足a1+a2+…+an=n^3
求an通项公式,求1/(a2-1)+1/(a3-1)+…+1/(a100-1)

Answer 1

简单分析一下，详情如图所示

Answer 2

a1=1
a1+a2+……+an=n^3【1】
a1+a2+……+a(n-1)=(n-1)^3【2】
【1】-【2】得an=n^3-(n-1)^3=3n²-3n+1
当n=1时也满足
于是an=3n²-3n+1
(2)
(an)-1=3n²-3n=3n(n-1)
1/[(an)-1]=1/[3n(n-1)]=(1/3)[1/(n-1)-1/n]
于是
1/(a2-1)+1/(a3-1)+…+1/(a100-1)
=(1/3)[1-(1/2)]+(1/3)[(1/2)-(1/3)]+……+(1/3)[(1/99)-(1/100)]
=(1/3)[1-(1/100)]
=33/100