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

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
a1=1

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]
再问： { n(n+1) - [∑(i:1->n) i .2^(i-1) ] } 是什么意思呢？
再答： 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+...+an
再问： S(n-1) = 2a(n-1) -2(n-1) +1 (2) (1)-(2) an = 2an- 2a(n-1) +2 an= 2a(n-1) -2 不应该是an = 2an- 2a(n-1) -2吗？
再答： an = 2an- 2a(n-1) +2 2an-an = 2a(n-1) -2 an = 2a(n-1) -2
再问： 应该An=3×2^（n-1）-2
再答： 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) =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) ] } 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) + 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]