Question

已知数列{an},其前n项和为Sn,且Sn=3/2n²+7/2n

Answer 1

an=Sn-S(n-1)

Sn=3/2n^2+7/2n

S(n-1)=3/2(n-1)^2+7/2(n-1)

Sn-S(n-1)=3/2n^2+7/2n-[3/2(n-1)^2+7/2(n-1)]

=3n-3/2+7/2

=3n+2

Bn=2^(3n+2)

B(n-1)=2^(3n+2-3)=2^(3n+2)*1/8

q=1/8

B1=2^5

Bn=32*(1/8)^(n-1)

前n项和Ta= 32*（1-1/8^n)/(1-1/8)=28(1-1/8^n)