Question

设数列{an}的前n项和为Sn，已知a1=a，a（n+1）=2Sn-2^n，n∈N*

设bn=Sn-2^n，求数列{bn}的通项公式

Answer 1

由a(n+1)=2Sn-2^n
知an=2S(n-1)-2^(n-1)
相减得a(n+1)-an=2an-2^(n-1)
a(n+1)=3an-2^(n-1)=3an-(3-2)×2^(n-1)=3[an-2^(n-1)]+2^n
则a(n+1)-2^n=3[an-2^(n-1)]
即数列{an-2^(n-1)}是以a1-2º=a-1为首项，3为公比的等比数列，设其前n项和为Tn
Tn=(a-1)(1-3^n)/(1-3)=(a-1)(3^n-1)/2
Sn=a1+a2+…+an
=(a1-2º)+(a2-2¹)+…+[an-2^(n-1)]+2º+2¹+…+2^(n-1)
=Tn+(1-2^n)/(1-2)
=[(a-1)(3^n-1)/2]+2^n-1
bn=[(a-1)(3^n-1)/2]-1=[(a-1)3^n-a-1]/2

Answer 2

bn=[(a-1)(3^n-1)/2]-1=[(a-1)3^n-a-1]/2