Question

高一数学问题求解：？

数列{an}各项均为正数，前n项和为Sn,首项为a1,且1/2,an,sn成等差数列，若an²=(1/2)^bn,设Cn=bn/an，那么{Cn}的前n项和为？

Answer 1

仅供参考：
1/2,an,sn成等差数列
∴1/2+Sn=2an
an=Sn-S(n-1)
1/2+Sn=2Sn-2S(n-1)
Sn=2S(n-1)+1/2
Sn+1/2=2(S(n-1)+1/2)
{Sn+1/2}是等比数列
∴Sn+1/2=2^(n-1)(S1+1/2)
1/2+S1=2a1 则a1=1/2
Sn=2^(n-1)-1/2
an=Sn-S(n-1)=2^(n-1)-2^(n-2)=2^(n-2)
an=2^(n-2)
an²=2^(2n-4)
则bn=4-2n
cn=(4-2n)/2^(n-2)=(4-2n)*(1/2)^(n-2)
c1=2*(1/2)^(-1)
c2=0*(1/2)^(0)
c3=-2*(1/2)^(1)
c4=-4*(1/2)^(2)
....
cn=(4-2n)*(1/2)^(n-2)=(2n-4)*(1/2)^(n-2)
n<=2时
{Cn}的前n项和=4
n>=3时(使用错位相减求和）
{Cn}的前n项和=4-2*(1/2)^(1)-4*(1/2)^(2)-6*(1/2)^(3)-.......-(2n-4)*(1/2)^(n-2)
=4-[2*(1/2)^(1)+4*(1/2)^(2)+6*(1/2)^(3)+.......+(2n-4)*(1/2)^(n-2)]
设Tn=2*(1/2)^(1)+4*(1/2)^(2)+6*(1/2)^(3)+......+(2n-4)*(1/2)^(n-2)
1/2Tn= 2*(1/2)^(2)+4*(1/2)^(3)+......+(2n-2)*(1/2)^(n-2)+(2n-4)*(1/2)^(n-1)

Tn-1/2Tn=1/2Tn=2*(1/2)^(1)+2*(1/2)^(2)+4*(1/2)^(3)+......+2*(1/2)^(n-2)-(2n-4)*(1/2)^(n-1)
=2*1/2*[1-(1/2)^(n-2)]/(1-1/2)-(2n-4)*(1/2)^(n-1)
=2[1-(1/2)^(n-2)]-(2n-4)*(1/2)^(n-1)
Tn=4[1-(1/2)^(n-2)]-2(2n-4)*(1/2)^(n-1)
=4-4(1/2)^(n-2)-(2n-4)*(1/2)^(n-2)
=4-2n*(1/2)^(n-2)
=4-n*(1/2)^(n-3)
∴n>=3时
{Cn}的前n项和=4-Tn=n*(1/2)^(n-3)
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Answer 2

Sn=2an-0.5 , Sn-1= 2an-1 -0.5
两式相减得：Sn - Sn-1 = an = 2an - 2an-1,即：an=2an-1
因为S1=a1=2a1-0.5 , 所以,a1=0.5
an=(1/4)*2^n = 2^(n-2)

an²=(1/2)^bn
bn= log(1/2)[2^(2n-4)]= 4-2n

Cn= (4-2n)/[2^(n-2)]
= 4/[2^(n-2)] - 2n/[2^(n-2)]
= (1/2)^(n-4)- n*(1/2)^(n-3)
其中，｛(1/2)^(n-4) ｝前n项之和是: 16- (1/2)^(n-4)
令Tn= {n*(1/2)^(n-3)}前n项之和
Tn - Tn /2
= [(1/2)^(-2) + 2*(1/2)^(-1) + ……+ n*(1/2)^(n-3) ] - [ (1/2)^(-3) + 2*(1/2)^(-2) + ……+ n*(1/2)^(n-4)]
= - (1/2)^(-3) - [ (1/2)^(-2) + (1/2)^(-1)+……+ (1/2)^(n-3)] + n*(1/2)^(n-3)
= - [16- (1/2)^(n-3)]+ n*(1/2)^(n-3)
Tn = -32+ (1/2)^(n-4)+ n*(1/2)^(n-4).
所以，｛Cn｝前n项和是: 16 + n*(1/2)^(n-4).

Answer 3

1/2,an,sn成等差数列,得，1/2+sn=2an=2(Sn-Sn-1),整理得，Sn=2Sn-1+1/2,可推出Sn-1=2Sn-2+1/2
两式相减，得an=2(Sn-1-Sn-2),即an=2an-1,an为等比数列，公比为2
an=a1*2^n-1
Cn=bn/an=bn/根号bn/2=2根号an
所以Cn=2根号a1*（根号2）^n-1
Cn也是等比数列
Cn的前n项和为：2根号下a1*(1-(根号2)^n)/1-根号2