设数列{an}的前n项和为sn,已知a1=a,an+1=sn+3^n,n∈N* (1)设bn=sn-3^n,求数列{bn}.Sn+1-3^n+1=2(Sn-3^n)怎么来的?
问题描述:
设数列{an}的前n项和为sn,已知a1=a,an+1=sn+3^n,n∈N* (1)设bn=sn-3^n,求数列{bn}.Sn+1-3^n+1=2(Sn-3^n)怎么来的?
答
Sn+1=an+1+Sn.
又an+1=Sn+3∧n.
Sn+1=2Sn+3∧n.①
3∧n+1=3×3∧n.②
①-②得
Sn+1-3∧n+1=2(Sn-3∧n)
后面的你知道吧,我就不说了.