Question

设｛an｝的各项都是正数的等比数列,bn=log1/3(下标)an

Answer 1

(1)设an=k*a(n-1)

则bn=log1/3 an=log1/3 [k*a(n-1)]=log1/3 k +log1/3 a(n-1);

所以bn-b(n-1)=log1/3 k;

(2)社b2=b1+d,b3=b1+2d带入

得b1=-1；b2=1;b3=3

所以bn=2n-3;

an=(1/3)^(2n-3)