已知an=2n+3n,bn=an+1+kan,若{bn}是等比数列,则k=______.
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已知an=2n+3n,bn=an+1+kan,若{bn}是等比数列,则k=______.
因为{bn} 是等比数列,故有
(an+1+kan)2=(an+2+kan+1)(an+kan-1),
将an=2n+3n代入上式,得
[2n+1+3n+1+k(2n+3n)]2
=[2n+2+3n+2+k(2n+1+3n+1)]•[2n+3n+k(2n-1+3n-1)],
即[(2+k)2n+(3+k)3n]2
=[(2+k)2n+1+(3+k)3n+1][(2+k)2n-1+(3+k)3n-1],
整理得
1
6(2+k)(3+k)•2n•3n=0,
解k-=2或k=-3.
故答案为:-2或-3
(an+1+kan)2=(an+2+kan+1)(an+kan-1),
将an=2n+3n代入上式,得
[2n+1+3n+1+k(2n+3n)]2
=[2n+2+3n+2+k(2n+1+3n+1)]•[2n+3n+k(2n-1+3n-1)],
即[(2+k)2n+(3+k)3n]2
=[(2+k)2n+1+(3+k)3n+1][(2+k)2n-1+(3+k)3n-1],
整理得
1
6(2+k)(3+k)•2n•3n=0,
解k-=2或k=-3.
故答案为:-2或-3
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