数列{an}满足递推式an=3an-1+3^n-1 (n>=2),其中a4=365 1.求a1,a2,a32.若存在一个实数λ,使得{(an+λ)/(3^n)为等差数列,求λ值;3.求数列{an}的前n项和.

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$数列{an}满足递推式an=3an-1+3^n-1 (n>=2),其中a4=365 1.求a1,a2,a32.若存在一个实数λ,使得{(an+λ)/(3^n)为等差数列,求λ值;3.求数列{an}的前n项和.$

数列{an}满足递推式an=3an-1+3^n-1 (n>=2),其中a4=365 1.求a1,a2,a32.若存在一个实数λ,使得{(an+λ)/(3^n)为等差数列,求λ值;3.求数列{an}的前n项和.
数列{an}满足递推式an=3an-1+3^n-1 (n>=2),其中a4=365 1.求a1,a2,a32.若存在一个实数λ,使得{(an+λ)/(3^n)为等差数列,求λ值;3.求数列{an}的前n项和.

数列{an}满足递推式an=3an-1+3^n-1 (n>=2),其中a4=365 1.求a1,a2,a32.若存在一个实数λ,使得{(an+λ)/(3^n)为等差数列,求λ值;3.求数列{an}的前n项和.
(1)a4=3a3 3^4-1=365 ∴a3=95 a3=3a2 3^3-1=95 ∴a2=29 a2=3a1 3^2-1=29 ∴a1=7 (2)设bn=(an λ)/3^n 要使其为等差数列则bn-b(n-1)为一个常数 bn-b(n-1) =(an λ)/3^n-[a(n-1) λ]/3^(n-1) 把an=3a(n-1) 3^n-1代入得：bn-b(n-1)=1-(1 2λ)/3^n λ是实数,不能是关于n的代数式 ∴1 2λ=0 ∴y=-1/2 (3)待定系数法：an-k=3[a(n-1)-k] 化简,得：an=3a(n-1)-2k=3a(n-1) 3^n-1 ∴k=(1-3^n)/2 ∴数列{an-(1-3^n)/2}是以8为首项,3为公比的等比数列根据等比数列求和公式求出Tn 之后Tn (1-3^1)/2 (1-3^2)/2 …… (1-3^n)/2即可得到数列{an}的前n项和Sn

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