Question

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

Answer 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]

追问

{ n(n+1) - [∑(i:1->n) i .2^(i-1) ] } 是什么意思呢？

追答

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

追问

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？

追答

an = 2an- 2a(n-1) +2

2an-an = 2a(n-1) -2
an = 2a(n-1) -2

追问

应该An=3×2^（n-1）-2

追答

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]

Answer 2

既然 Sn = 2An - 2n + 1，
则 S(n-1)= 2A(n-1) - 2(n-1) +1
所以有：An = Sn - S(n-1) = (2An -2n +1) - [2A(n-1) - 2(n-1) +1] = 2An - 2A(n-1) +2
化简可得：
An = 2A(n-1) - 2
An - 2 = 2A(n-1) - 2 - 2 = 2[A(n-1) - 2]
即 ｛An - 2｝是一个等比数列，因此有：
An - 2 = (A1-2) * 2^(n-1) = -1* 2^(n-1)，即 An = 2 - 2^(n-1)

新数列{nAn/3} 的通项为：
nAn/3 = 2n/3 - 2^(n-1)/3
所以，Tn = 2/3*(1+2+3+ ……+n) - 1/3* [2^0 + 2^1 + …… + 2^(n-1)]
= 2/3 * n(n+1)/2 -1/3 * (2^n-1)
= 1/3*[n(n+1) - (2^n -1)]

Answer 3

A1＝2A1-2+1知A1＝1,Sn-Sn-1＝An=2An-2An-1+2,知An＝2An-1-2,知An=3×2∧（n-1）-2
nAn/3=n×2∧n-1）-2n/3
Tn=（n-1）2∧n-（n²+n）/3+1

Answer 4

令n=n-1求出通项