Question

已知an是各项均为正数的等比数列

a1+a2=2(a1分之1+a2分之1)，a2+a3+a4=64（a2分之1+a3分之1+a4分之1）求an的通项公式 谢谢

Answer 1

等比数列的公比为q
a1+a1q=2(1/a1+1/a1q)
(1+q)a1^2=2(1+1/q)
(1+q)a1^2=2(1+q)/q
a1^2=2/q (1)
a2+a3+a4=64（a2分之1+a3分之1+a4分之1）
a1q+a1q^2+a1q^3=64(1/a1q+/a1q^2+1/a1q^3)
q(1+q+q^2)a1=64(1/a1q+/a1q^2+1/a1q^3)
q(1+q+q^2)a1^2=64(1+/q+1/q^2)/q
q(1+q+q^2)a1^2=64(1+q+q^2)/q^3
a1^2=64/q^4 (2)
2/q=64/q^4
q^3=32
q=2^(5/3)
a1^2=2/q
a1^2=2^(-2/3)
a1=2^(-1/3)
an=a1q^(n-1)
an=2^(-1/3)(2^(5/3))^(n-1)
an=2^(-1/3+5(n-1)/3)
an=2^(5n-6/3)