已知数列{An}满足A1=1,A2=3,A(n+2)=3A(n+1)-2An,求an的通项公式

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已知数列{An}满足A1=1,A2=3,A(n+2)=3A(n+1)-2An,求an的通项公式
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已知数列{An}满足A1=1,A2=3,A(n+2)=3A(n+1)-2An,求an的通项公式
已知数列{An}满足A1=1,A2=3,A(n+2)=3A(n+1)-2An,求an的通项公式

已知数列{An}满足A1=1,A2=3,A(n+2)=3A(n+1)-2An,求an的通项公式
a1=1,a2=3,a(n+2)=3a(n+1)-2an.
法一:待定系数法.
设待定系数s、t,使a(n+2)-sa(n+1)=t(a(n+1)-san).
整理得a(n+2)=(s+t)a(n+1)-stan.
对比原式,得s+t=3,st=2.
解得s=1,t=2或s=2,t=1.
用后一组解,有a(n+2)-2a(n+1)=a(n+1)-2an,a2-2a1=1.
∴数列{a(n+1)-2an}是首项a2-2a1=1,公比q=1的等比数列.
∴a(n+1)-2an=a2-2a1=1,故a(n+1)=2an+1.
则a(n+1)+1=2(an+1),a1+1=2.
∴数列{an+1}是首项a1+1=2,公比q=2的等比数列.
∴an+1=(a1+1)×qⁿ⁻¹=2ⁿ
∴an=2ⁿ-1.
综上,数列{an}的通项公式为an=2ⁿ-1.
法二:数学归纳法.
a1=1,a2=3.
猜想an=2ⁿ-1.
①当n=1、2时,猜想显然成立.
②假设当n=k、k+1时结论成立,则有ak=2^k-1,a(k+1)=2^(k+1)-1.
③当n=k+2时:
a(k+2)=3a(k+1)-2ak
=3×2^(k+1)-3-2×2^k+2
=2×2^(k+1)-1
=2^(k+2)-1.
显然,n=k+2时结论也成立.
综上,由①、②、③得对任意n∈N*,an=2ⁿ-1.
法三:特征方程法.
a(n+2)=3a(n+1)-2an
其特征方程为x^2=3x-2,解得x1=1,x2=2.
从而an=c₁x1ⁿ+c₂x2ⁿ=c₁+c₂×2ⁿ.
代入a1、a2的值,得c1+2c2=1,c1+4c2=3.
解得c1=-1,c2=1,故an=2ⁿ-1.
综上,数列{an}的通项公式为an=2ⁿ-1.

a(n+2) = 3a(n+1) -2an
a(n+2) - 2a(n+1) = a(n+1) -2an
= a2- 2a1
=1
an - 2a(n-1) =1
an +1 = 2(a(n-1) +1 )
= 2^(...

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a(n+2) = 3a(n+1) -2an
a(n+2) - 2a(n+1) = a(n+1) -2an
= a2- 2a1
=1
an - 2a(n-1) =1
an +1 = 2(a(n-1) +1 )
= 2^(n-2) . ( a2 +1)
= 2^n
an = -1+ 2^n

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