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.
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![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.](/uploads/image/z/214609-49-9.jpg?t=19.%E4%B8%80%E7%9B%B4%E6%95%B0%E5%88%97An%2CA1%3Dm%2CA%28n%2B1%29%3D2An%2B3%5E%28n%2B1%29.%281%29%E8%AE%BEBn%3DA%28n%2B1%29%2F3%5En%2C%E6%B1%82Bn%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F.)
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.
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.
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.
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.
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][...
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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.
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