Question

已知数列an的前n项和为sn满足Sn=2an-3n（n属于正整数）.1求证an+3为等比数列并求an的通项公式.2数列a

Answer 1

1.

Sn=2An-3n

S(n-1)=2A(n-1)-3(n-1)

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

An=2A(n-1)+3

An+3=2A(n-1)+6=2(A(n-1)+3)

A1=S1=2A1-3

A1=3

A1+3=3+3=6

{An+3}是以6为首项,2为公比的等比数列

An+3=6×2^(n-1)

An=3×2^n-3

2.

很明显{An}是递增数列

假设Ax,Ay,Az三项成等差,0