Question

若an与an*2/n均为等差数列,且a1=1/2

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=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)