Question

已知对于任意正整数n都有a1+a2+...+an=n^3,则(1/a2-1)+(1/a3-1)+...+(1/a100-1)=_____

Answer 1

a1+a2+...+a(n-1)+an=n³ (1)
a1+a2+...+a(n-1)=(n-1)³ (2)
(1)-(2)
an=n³-(n-1)³
=[n-(n-1)][n²+n(n-1)+(n-1)²]
=3n²-3n+1
1/(an -1)=1/(3n²-3n+1-1)=1/(3n²-3n)=(1/3)×1/(n²-n)=(1/3)×1/[n(n-1)]=(1/3)[1/(n-1)-1/n]

1/(a2-1)+1/(a3-1)+...+1/(a100 -1)
=(1/3)[1/1 -1/2+1/2-1/3+...+1/(99)-1/100]
=(1/3)(1 -1/100)
=(1/3)(99)/100
=33/100

Answer 2

Sn=n^3 (1)
S(n-1) =(n-1)^3 (2)
(1)-(2)
an= 3n^2-3n+1
an-1= 3n(n-1)
1/(an-1) = (1/3)( 1/(n-1) -1/n )
(1/(a2-1))+(1/(a3-1))+...+(1/(a100-1) )

= (1/3)( 1- 1/100)
= 33/100