第1个回答 2011-02-05
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.