Question

已知Sn为数列An的前n项和,且Sn=2-2An 1.求证数列 An为等比数列

已知Sn为数列An的前n项和,且Sn=2-2An 1.求证数列 An为等比数列 2.求数列An的通项公式 3.求数列 ｛AnSn｝的前n项和Wn
要具体的过程

Answer 1

an=Sn-S(n-1)=-2an+2a(n-1)
3an=2a(n-1)
an/a(n-1)=2/3
等比数列，公比=2/3

S1=2-2a1=a1,
a1=2/3
an=a1*q^(n-1)=(2/3)^n

anSn=(2/3)^n *(2-2*(2/3)^n)
=2(2/3)^n-2(2/3)^(2n)
=(4/3)(2/3)^(n-1)-(8/9)(4/9)^(n-1)
Wn=[(4/3)((2/3)^n-1)/((2/3)-1)]-[(8/9)((4/9)^n-1)/((4/9)-1)]
=4-4(2/3)^n+(8/5)(4/9)^n-(8/5)
=(12/5)-4(2/3)^n+(8/5)(4/9)^n

Answer 2

Sn=2-2An
Sn-1=2-2An-1
An=-2An+2An-1
An=2/3(An-1)
A1=2-2A1 A1=2/3
An=(2/3)^n