Question

高中数学数列，通项公式问题

19.一直数列An,A1=m,A(n+1)=2An+3^(n+1).
(1)设Bn=A(n+1)/3^n,求Bn通项公式。

Answer 1

A(n+1)=2An+(3-2)*3^(n+1), A(n+1)-3^(n+2)=2(An-3^(n+1)),
令Cn=An-3^(n+1)，则C1=m-9，Cn=(m-9)*2^(n-1).
故An=(m-9)*2^(n-1)+3^(n+1)。于是Bn=(m-9)*(2/3)^n+9.

Answer 2

将A(n+1)=2An+3^(n+1),两边同时除以2^(n+1),得A(n+1)/(2^(n+1))-An/2^n=(3/2)^(n+1),An/(2^n)-A(n-1)/2^(n-1)=(3/2)^n......A2/(2^2)-A1/2=(3/2)^2,n项累加得：A(n+1)/(2^(n+1))-A1/2=(3/2)^2+(3/2)^3+...+(3/2)^(n+1)=[(3/2)^2][1-(3/2)^n]/[1-(3/2)]=-2[(3/2)^2-(3/2)^(n+2)]注意到A1=m,移项整理得：A(n+1)=(m-9)2^n+3^(n+2),于是Bn=A(n+1)/3^n=(m-9)(2/3)^n+9.