Question

等比数列{an}各项为正数，Sn是其前n项和，且a1+a5=34，a2a4=64，求{an}的公比q和Sn

Answer 1

解:
a1+a5=a1+a1q^4=34①

a1^2q^4=64②
联立①②
解得a1=2
q=2
所以Sn=[2(1-2^2)]/(1-2)

Answer 2

a1+a5
=a1+a1q^4=34
a1^2+a1^2q^4=34a1
a1^2q^4=34a1-a1^2
a2a4
=a1q*a1q^3=a1^2q^4=64 1
34a1-a1^2=64
a1^2-34a1+64=0
(a1-32)(a1-2)=0
a1=32 或a1=2
分别代入1式得
q=1/2 或 q=2
当q=1/2时
Sn=a1(1-q^n)/(1-q)
=32(1-1/2^n)/(1-1/2)
=64(1-1/2^n)
=64-2^(6-n)

当q=2时
Sn=a1(q^n-1)/(q-1)
=2(2^n-1)/(2-1)
=2^(n+1)-2

Answer 3

a1 a5=34 a2a4=64 a3的平方等于64 所以a3=8 a1=a3/q^2=8/q^2 a5=a3q^2=8q^2 所以8/q^2 8q^2=34 整理得8 8q^4=34q^2 令x=q^2 则8 8x^2=34x 解得x=4或1/4 q=2或1/2 a1=2或32 sn=2^(n 1)-2或64-1/2^(n-6)

Answer 4

a5=a1*q4 a4=a1*q3
等比数列{an}各项为正数 q>0
最用sn的公式代入解