Question

已知等比数列{an}中，公比q属于(0,1)，a2+a4=5/4，a1a5=1/4，设bn=1/2nan。求an的通项公式，求数列bn的前n

Answer 1

等比数列,则有a2*a4=a1a5=1/4,又a2+a4=5/4
解得:a2=1,a4=1/4或a2=1/4,a4=1
(1)a2=1,a4=1/4,q^2=a4/a1=1/4,因0<q<1,则有q=1/2
(2)a2=1/4,a4=1.q^2=a4/a2=4,不符合,舍.
故q=1/2.
a1=a2/q=1/(1/2)=2
an=a1q^(n-1)=2*(1/2)^(n-1)
bn=1/2nan=1/2n*2*(1/2)^(n-1)=n*(1/2)^(n-1)=2n*(1/2)^n
和Sn=2[1*1/2+2*(1/2)^2+....+n(1/2)^n]
1/2Sn=2[1*(1/2)^2+2*(1/2)^3+...+n(1/2)^(n+1)]
Sn-1/2Sn=2[1/2+(1/2)^2+(1/2)^3+...+(1/2)^n-n*(1/2)^(n+1)]
1/2Sn=2{(1/2)*[1-(1/2)^n]/(1-1/2)-n*(1/2)^(n+1)}
Sn=4[1-(1/2)^n-n*(1/2)^(n+1)]

Answer 2

a2+a4=5/4 …, a1*a5=a2*a4=1/4
(a2-a4)2=(a2+a4)2-4a2*a4=9/16
∵q∈(0,1)
∴a2-a4=3/4 ,与联立得
∴a2=1，a4=1/4
q2=a4/a2=1/4
∴q=1/2
∴an=2*(1/2)(n-1)=22-n
bn=n*21-n
b1=1*20，b2=2*(1/2)1 b3=3*(1/2)2…
Sn=1*20+2*(1/2)1 +3*(1/2)2 +…+ n*21-n
（1/2）Sn= 1*(1/2)1 +2*(1/2)2 +…+(n-1)*21-n+n*2-n
错位相减，Sn=4+n/2n-1-(1/2)n-2

Answer 3

a2+a4=a1q+a1qqq=a1q(1+qq)=5\4,a1a5=a1a1qqqq=1\4,然后互相代入公式就可以求出a1 ,q
an=a1q（n-1）次方