设数列An的前n项和为Sn,已知a1=1,An+1=Sn+3n+1求证数列{An+3}是等比数列
问题描述:
设数列An的前n项和为Sn,已知a1=1,An+1=Sn+3n+1求证数列{An+3}是等比数列
答
证明:A(n+1)=Sn+3n+1,则An=S(n-1)+3n-2
两式想减得A(n+1)-An=Sn+3n+1-(S(n-1)+3n-2)=An+3
即A(n+1)+3=2(An+3)
即(A(n+1)+3)/(An+3)=2
又a1=1 A1+3=4
即{An+3}是首项为4,公比为2的等比数列