已知数列An的前n项和为Sn,且An+2Sn=4N(N∈N+).1) 求数列An的通项公式 2

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已知数列An的前n项和为Sn,且An+2Sn=4N(N∈N+).1) 求数列An的通项公式 2
已知数列An的前n项和为Sn,且An+2Sn=4N(N∈N+).
1) 求数列An的通项公式
2) 若Bn=NAn,求数列Bn的前n项和Tn.

(1)
an +2Sn= 4n
n=1 ,a1= 4/3
2Sn = 4n - an (1)
2S(n-1) = 4(n-1) - a(n-1) (2)
(1)-(2)
2an = 4 - an + a(n-1)
3an= a(n-1) + 4
3(an-2) = a(n-1) -2
an - 2 = (1/3) [ a(n-1) -2 ]
{an - 2} 是等比数列,q= 1/3
an - 2 = (1/3)^(n-1) .(a1 - 2)
= -2(1/3)^n
an = 2-2(1/3)^n
(2)
bn = nan
= 2n - 2n(1/3)^n
Tn = b1+b2+...+bn
= n(n+1) - 2[∑(i:1->n) i.(1/3)^i ]
let
S = 1.(1/3)^1 + 2.(1/3)^2+...+n.(1/3)^n (1)
(1/3)S = 1.(1/3)^2 + 2.(1/3)^3+...+n.(1/3)^(n+1) (2)
(1)-(2)
(2/3)S = [1/3^1 + 1/3^2+...+1/3^n] - n(1/3)^(n+1)
= (1/2)(1-1/3^n) -n(1/3)^(n+1)
S = (3/2) [(1/2)(1-1/3^n) -n(1/3)^(n+1)]
Tn = n(n+1) - 2S
= n(n+1) - 3[(1/2)(1-1/3^n) -n(1/3)^(n+1)]

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