在等比数列中,a2=1/4,a3*a6=1/512.求an设bn=log以二an^2为底2的对数*log以二an+1^2为底2的对数，求前n项和tn

Question

在等比数列中,a2=1/4,a3*a6=1/512.求an
设bn=log以二an^2为底2的对数*log以二an+1^2为底2的对数，求前n项和tn

Answer 1

a3a6=a2q*a2q^4=1/16 q^5=1/512
得：q^5=1/32
得q=1/2
故an=a2q^(n-2)=1/4(1/2)^(n-2)=1/2^n

2an²=2/2^(2n)=2^(1-2n)
2a(n-1)²=2^(3-2n)
bn=log2an²(2)* log2a(n+1)²(2)=1/(1-2n)*1/(3-2n)=1/[(2n-1)(2n-3)]=0.5[1/(2n-3)-1/(2n-1)]
因此tn=0.5[-1-1+1-1/3+1/3-1/5+...+1/(2n-3)-1/(2n-1)]=0.5[-1-1/(2n-1)]=-n/(2n-1)