Question

数列an的前n项和Sn满足：Sn=2n-an 求通项公式

Answer 1

Sn=2n-An

Sn+1=2(n+1)-A(n+1)

相减得

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

2A（n+1）=An+2

2A(n+1)-4=An-2

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

设Bn=An-2

所以Bn是等比数列,公比为1/2

B1=A1-2=-1

Bn=B1*(1/2)^(n-1)

Bn=-(1/2)^(n-1)

An=Bn+2=2-(1/2)^(n-1)