Question

这题怎么做求解，谢谢！

Answer 1

18. 记 a1 = b1 = c, 则 a3 = c+2d, a5 = c+4d, b3 = cd^2, b5 = cd^4
则 c+2d = 3cd^2, c+4d = 5cd^4
2d = c(3d^2-1), 4d = c(5d^4-1)
两式相除得 2 = (5d^4-1)/(3d^2-1)， 5d^4-6d^2+1 = 0,
d > 0, d ≠ 1，得 d = 1/√5
c = -√5, {an} = (n-6)/√5, (bn} = -5/5^(n/2).

Answer 2

a3=a1+2d

a5=a1+4d
b3=b1*d^2

b5=b1*d^4
a1+2d=b1*d^2

a1+4d=5b1*d^4
2d=3a1*d^2-a1 (1)
4d=a1*d^4-a1 (2)
(2)/(1)
2=(5d^4-1)/(3d^2-1)
5d^4-6d^2+1=0
d^2=1 (舍去) d^2=1/5
d=√5/5 a1=-√5
an=-√5+(n-1)*√5/5
bn=-√5*(√5/5)^(n-1)