n=a(n+1)-an ,b(n+1)-bn=6
(1)
Solution:
b(n+1)-bn=6
bn-b1 = 6(n-1)
bn = 6(n-1) +b1
bn=a(n+1)-an
a(n+1)-an = 6(n-1) +b1
an-a(n-1) = 6(n-2) +b1
an - a1 = 6[0+1+...+(n-2) ] + (n-1)b1
= 3(n-2)(n-1)+ (n-1)b1
an =3(n-2)(n-1)+ (n-1)b1 +a1
(2)
a1=-b1=a
min{an} =a6
To find :range of a
an =3(n-2)(n-1)+ (n-1)b1 +a1
=3(n-2)(n-1)- (n-1)a +a
=3n^2-(9+a)n+ (6+2a)
let
f(x) = 3x^2-(9+a)x+ (6+2a)
f'(x) = 6x- (9+a)
f'(5) = 30- (9+a) 21
f'(7)= 42- (9+a) >0
a< 33
ie
21