已知{an}是各项均为正数的等比数列,且a1+a2=2(1/a1+1/a2),a3+a4+a5=64(1/a3+1/a4+1/a5)求{an}的通项公式

设an=a1*q^(n-1),a1＞0，q＞0
a2=a1q
a3=a1q^2
a4=a1q^3
a5=a1q^4
1/a2=1/(a1q)
1/a3=1/(a1q^2)
1/a4=1/(a1q^3)
1/a5=1/(a1q^4)
a1+a2-2(1/a1+1/a2)=a1+a1q-2[1/a1+1/(a1q)]
=[a1-2/(a1q)](1+q)
=0
a1-2/(a1q)=0
a1^2*q=2
a3+a4+a5-64(1/a3+1/a4+1/a5)=a1q^2+a1q^3+a1q^4-64[1/(a1q^2)+1/(a1q^3)+1/(a1q^4)]
=a1q^2(1+q+q^2)-64[1/(a1q^4)](1+q+q^2)
=(1+q+q^2)a1[q^2-64/(a1^2*q^4)]
=(1+q+q^2)a1[q+8/(a1*q^2)][q-8/(a1*q^2)]
=0
q-8/(a1*q^2)=0
a1*q^3=8
a1=1,q=2
an=2^(n-1)

设公比为q
a1+a2=2(1/a1+1/a2)=>a1(1+q)=(2/a1q)*(q+1)=>a1^2*q=2
a3+a4+a5=64(1/a3+1/a4+1/a5)=>a3(q^2+q+1)=64/(a3*q^2)(q^2+q+1)=>(a3*q)^2=a1^2*q^6=64
因为{an}各项均为正数,所以a4=a3*q=8
而q^5=64/2=32,q=2
所以a1=1,an=2^(n-1)