
已知数列{an}的前n项和为Sn,且Sn=1-1/2an(n∈N*),已知数列{bn}的通项公式bn=
已知数列{bn}的通项公式bn=2n-1,记cn=an*bn,求数列{cn}的前n项和Tn。补充一下:这是第二问,第一问{an}的通项公式是an=2/3*(1/3)^(n...
已知数列{bn}的通项公式bn=2n-1,记cn=an*bn,求数列{cn}的前n项和Tn。
补充一下:这是第二问,第一问{an}的通项公式是an=2/3 *(1/3)^(n-1). 展开
补充一下:这是第二问,第一问{an}的通项公式是an=2/3 *(1/3)^(n-1). 展开
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Sn=1-(1/2)an
n=1, a1= 2/3
an= Sn- S(n-1)
= -(1/2)an + (1/2)a(n-1)
an/a(n-1)=1/3
an/a1= (1/3)^(n-1)
an=2.(1/3)^n
bn=2n-1
cn =an.bn
=[2.(1/3)^n]. (2n-1)
= 4(n.(1/3)^n] - 2(1/3)^n
Tn =c1+c2+...+cn
= 4[∑(i:1->n) i.(1/3)^i] - (1- (1/3)^n )
let
S = 1.(1/3)+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)+(1/3)^2+...+(1/3)^n] - n.(1/3)^(n+1)
= (1/2)[1- (1/3)^n] - n.(1/3)^(n+1)
S = (3/4)[1- (1/3)^n] - (1/2)n(1/3)^n
Tn =c1+c2+...+cn
= 4[∑(i:1->n) i.(1/3)^i] - (1- (1/3)^n )
=4S - (1- (1/3)^n )
=3[1- (1/3)^n] - 2n(1/3)^n - (1- (1/3)^n )
= 2 -2(n+1).(1/3)^n
n=1, a1= 2/3
an= Sn- S(n-1)
= -(1/2)an + (1/2)a(n-1)
an/a(n-1)=1/3
an/a1= (1/3)^(n-1)
an=2.(1/3)^n
bn=2n-1
cn =an.bn
=[2.(1/3)^n]. (2n-1)
= 4(n.(1/3)^n] - 2(1/3)^n
Tn =c1+c2+...+cn
= 4[∑(i:1->n) i.(1/3)^i] - (1- (1/3)^n )
let
S = 1.(1/3)+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)+(1/3)^2+...+(1/3)^n] - n.(1/3)^(n+1)
= (1/2)[1- (1/3)^n] - n.(1/3)^(n+1)
S = (3/4)[1- (1/3)^n] - (1/2)n(1/3)^n
Tn =c1+c2+...+cn
= 4[∑(i:1->n) i.(1/3)^i] - (1- (1/3)^n )
=4S - (1- (1/3)^n )
=3[1- (1/3)^n] - 2n(1/3)^n - (1- (1/3)^n )
= 2 -2(n+1).(1/3)^n
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