已知数列{an}和{bn}中,a1=-10,b1=-13,且a下标n+1=-2a下标n+4b下标n,b下标n+1=-5a下标n+7b下标n,求an和bn的通项公式

Question

已知数列{an}和{bn}中,a1=-10,b1=-13,且a下标n+1=-2a下标n+4b下标n,b下标n+1=-5a下标n+7b下标n,求an和bn的通项公式

Answer 1

a(n+1)=-2an +4bn (1)
b(n+1) = -5an+7bn (2)
(1)-(2)
a(n+1)-b(n+1) = 3(an - bn)
{ an -bn } 是等比数列,q=3
an -bn = (a1-b1).3^(n-1)
= 3^n
bn = an - 3^n (3)
sub (3) into (1)
a(n+1) = -2an + 4an - 4.3^n
a(n+1) +4.3^(n+1) = 2[ an +4.3^n]
{an +4(3^n) } 是等比数列,q=2
an +4(3^n) = (a1+ 12).2^(n-1)
= 2^n
an = 2^n - 4(3^n)
from (3)
bn = an -3^n
= 2^n - 5(3^n)