Question

已知数列｛an｝的前n项和Sn,a1=2且4S（n+1）=3Sn+1,求

Answer 1

4S（n+1）=3Sn+1

4S(n+1)-3Sn=1

S(n+1)-3(S(n+1)-Sn)=1

S(n+1)-3A(n+1)=1

S(n+1)=3A(n+1)+1 1

Sn=3An+1 2

1-2得

S(n+1)-Sn=3A(n+1)+1-3An-1=3A(n+1)-3An=A(n+1)

2A(n+1)=3An

A(n+1)/An=3/2

所以是以公比3/2的等比数列

则

An=a1q^(n-1)

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

=2*(3/2)^n/ 3/2

=4/3 * (3/2)^n

Sn=3An+1

=3* 4/3 *(3/2)^n +1

=4*(3/2)^n+1