Question

: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=32(1/a3+1/a4) 求{an}的通项公式 描述:

Answer 1

a(n)=aq^(n-1), a>0, q>0.
a(1)+a(2)=a+aq=a(1+q)=2[1/a(1)+1/a(2)]=2[1/a + 1/(aq)]=[2/(aq)][q+1],
a=2/(aq), a^2q=2,
a(3)+a(4)=aq^2+aq^3=aq^2(1+q)=32[1/a(3)+1/a(4)]=32[1/(aq^2)+1/(aq^3)]=[32/(aq^3)][q+1],
aq^2=32/(aq^3), a^2q^5=32,
32/2 = 16=a^2q^5/(a^2q)=q^4, q=2.
2=a^2q=a^2, a=1.
a(n)=aq^(n-1)=2^(n-1)

Answer 2

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)