a(1)=k,
a(2)=(2/3)a(1)+1-4=2k/3-3=(2k-9)/3,
a(3)=(2/3)a(2)+2-4=(2/3)(2k-9)/3-2 = [4k-36]/9
若[a(2)]^2=a(1)a(3),則
(2k-9)^2/9=k(4k-36)/9,
(2k-9)^2=k(4k-36),
4k^2-36k+81=4k^2-36k,
81=0矛盾.
因此,[a(2)]^2不等于a(1)a(3),{a(n)}不是等比數(shù)列.
a(n+1)=(2/3)a(n)+n-4,
a(n+1)+x(n+1)+y=(2/3)a(n)+n-4+xn+x+y=(2/3)a(n)+(x+1)n+x+y-4
=(2/3)[a(n)+3(x+1)/2*n + 3(x+y-4)/2]
x=3(x+1)/2,x=-3,
y=3(x+y-4)/2,y=12-3x=21.
a(n+1)-3(n+1)+21=(2/3)a(n)+n-4-3(n+1)+21=(2/3)a(n)-2n+14=(2/3)[a(n)-3n+21]
{a(n)-3n+21}是首項(xiàng)為a(1)-3+21=k+18,公比為2/3的等比數(shù)列.
a(n)-3n+21=(k+18)(2/3)^(n-1)
b(n)=(-1)^n[a(n)-3n+21]=(-1)^n*(k+18)(2/3)^(n-1)=(-k-18)(-2/3)^(n-1)
-k-18不為0,也即k不等于-18時(shí),
{b(n)}是首項(xiàng)為(-k-18),公比為(-2/3)的等比數(shù)列.