4个回答
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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=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]
更多追问追答
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
展开全部
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)]
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)]
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展开全部
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
nAn/3=n×2∧n-1)-2n/3
Tn=(n-1)2∧n-(n²+n)/3+1
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