数列满足a0=1/4,及对于自然数n,a(n+1)=(an)^2+an,bn=1/(an+1),Sn为{bn}的前n项和.则S2011的整数部分为?

Question

数列满足a0=1/4,及对于自然数n,a(n+1)=(an)^2+an,bn=1/(an+1),Sn为{bn}的前n项和.则S2011的整数部分为?

Answer 1

a(n+1)=(an)^2+an
1/a(n+1) = 1/an(an+1)
1/(an+1) =1/an - 1/a(n+1)
Sn = (1/a0-1/a1) +(1/a1-1/a2)+...+(1/an-1/a(n+1))
=1/a0 -1/a(n+1)
= 4- 1/a(n+1)
S2011 =4- 1/a2012
S2011的整数部分=3