已知数列{an}是首项a1=1/4的等比数列,其前n项和sn中s3=3/16,通项公式an=（-1/2）^n+1设bn=log1/2(|an|)(以1/2为底）,Tn=1/b1b2+1/b2b3+.+1/bn+bn+1,求Tn.

Question

已知数列{an}是首项a1=1/4的等比数列,其前n项和sn中s3=3/16,通项公式an=（-1/2）^n+1
设bn=log1/2(|an|)(以1/2为底）,Tn=1/b1b2+1/b2b3+.+1/bn+bn+1,求Tn.

Answer 1

an = (-1/2)^(n+1)
bn = log |an|
= (n+1)
1/[bn.b(n+1)] = 1/[(n+1)(n+2)] =1/(n+1) -1/(n+2)
Tn=1/(b1b2)+1/(b2b3)+.+1/[bn.b(n+1)]
= 1/2 - 1/(n+2)