Question

数列{an}是正项等比数列,若a2=1-a1,a4=9-a3，则a4+a5=

Answer 1

公比为q q>0
a2=1-a1=a1q
a1(1+q)=1 (1)
a4=9-a3
a1*q^3=9-a1*q^2
a1*q^2(1+q)=9 (2)
(2)/(1) a1*q^2(1+q)/a1*(1+q)=9/1
q^2=9
q=3
a4+a5=a1q^3+a1q^4=a1q^3(1+q)=q^3=3^3=27

Answer 2

设等比数列{ a(n) }的公比为q
a(2)=1-a(1) => a(1)×(q+1)=1
a(4)=9-a(3) => a(1)×(q+1)×q²=9
所以：q²=9，考虑到所有a(n)>0，有：q=3
所以：a(5)+a(4) = a(1)×(q+1)×q³ = 3³ = 27

Answer 3

a1q=1-a1
a1(q+1)=1
a4=9-a3
a1q^3=9-a1q^2
a1(q^3+q^2)=9
a1q^2(q+1)=9
a1(q+1)=1
q^2=9
q=3
a1(q+1)=1
a1=1/4
a4+a5
=a1q^3+a1q^4
=a1q^3(1+q)
=1/4*27*4
=27