设数列An的前n项和Sn,满足Sn=2an-2n+1,n属于N 求数列{nAn/3}的前n项和Tn

Sn=2an-2n+1 (1)
n=1,a1= 1
S(n-1) = 2a(n-1) -2(n-1) +1 (2)
(1)-(2)
an = 2an- 2a(n-1) +2
an= 2a(n-1) -2
an -2 = 2[a(n-1) -2 ]
{an - 2 } 是等比数列,q=2
an - 2 = 2^(n-1) .(a1 - 2)
=-2^(n-1)
an = 2-2^(n-1)
nan/3 = (1/3)[2n - n.2^(n-1) ]
Tn = (1/3) { n(n+1) - [∑(i:1->n) i .2^(i-1) ] }
let
S = 1.2^0+2.2^1+...+n.2^(n-1) (1)
2S = 1.2^2+2.2^2+...+n.2^n (2)
(2)-(1)
S = n.2^n - [ 1+2+...+2^(n-1)]
=n.2^n - (2^n -1)
= 1+ (n-1).2^n
Tn = (1/3) { n(n+1) - [∑(i:1->n) i .2^(i-1) ] }
=(1/3)[ n(n+1) - S]
=(1/3)[ n^2n+n-1 - (n-1).2^n]{ n(n+1) - [∑(i:1->n) i .2^(i-1) ] } 是什么意思呢？nan/3 = (1/3)[2n - n.2^(n-1) ]Tn = (1/3)∑(i:1->n) (iai)= (1/3) {∑(i:1->n)[2i - i.2^(i-1)] }=(1/3) { n(n+1) - [∑(i:1->n) i .2^(i-1) ] } ∑(i:1->n) i = 1+2+3+...+ n∑(i:1->n) ai = a1+a2+...+anS(n-1) = 2a(n-1) -2(n-1) +1(2)(1)-(2)an = 2an- 2a(n-1) +2an= 2a(n-1) -2不应该是an = 2an- 2a(n-1) -2吗？an = 2an- 2a(n-1) +22an-an = 2a(n-1) -2an = 2a(n-1) -2应该An=3×2^（n-1）-2Sn=2an-2n+1 (1)n=1, a1= 1S(n-1) = 2a(n-1) -2(n-1) +1(2)(1)-(2)an = 2an- 2a(n-1) -2an= 2a(n-1) +2an +2 = 2[a(n-1) +2 ]{an + 2 } 是等比数列,q=2an + 2 = 2^(n-1) .(a1 + 2) =3.2^(n-1)an = -2+3.2^(n-1)nan/3 = (1/3)[-2n + 3n.2^(n-1) ]Tn = (1/3) { -n(n+1) - 3[∑(i:1->n) i .2^(i-1) ] } letS = 1.2^0+2.2^1+...+n.2^(n-1)(1)2S = 1.2^2+2.2^2+...+n.2^n(2)(2)-(1)S = n.2^n - [ 1+2+...+2^(n-1)]=n.2^n - (2^n -1)= 1+ (n-1).2^nTn = (1/3) { -n(n+1) + 3[∑(i:1->n) i .2^(i-1) ] } =(1/3)[ -n(n+1) + 3S] =(1/3)[ -n^2-n+3 + 3(n-1).2^n]