已知数列{an}的前n项和Sn,且Sn+1/2an=1设bn=log3（1-Sn+1）,求适合方程1/b1b2+1/b2b3+``````+1/bnbn+1=25/51的n的值

Question

已知数列{an}的前n项和Sn,且Sn+1/2an=1
设bn=log3（1-Sn+1）,求适合方程1/b1b2+1/b2b3+``````+1/bnbn+1=25/51的n的值

Answer 1

Sn+(1/2)an=1
n=1,a1= 2/3
Sn+(1/2)an=1
Sn+(1/2)[Sn-S(n-1)]=1
(3/2)Sn = (1/2)S(n-1) +1
Sn = (1/3)S(n-1) + (2/3)
Sn - 1 = (1/3)(S(n-1) - 1)
{Sn - 1} 是等比数列,q=1/3
Sn - 1 = (1/3)^(n-1) .[S1 - 1]
=-1/3^n
Sn = 1- 1/3^n
bn=log[1-S(n+1) ]
=log[1/3^(n+1) ]
=-(n+1)
1/[bn.b(n+1)] = 1/[(n+1)(n+2)] = 1/(n+1) - 1/(n+2)
1/(b1b2)+1/(b2b3)+...+1/(bnb(n+1)) = 25/51
1/2 - 1/(n+2) = 25/51
1/(n+2) = 1/102
n+2= 102
n=100