Question

已知a﹤b,且a^2-a-6=0,b^2-b-6=0,数列{an}{bn}满足a1=1,a2=-6a,an+1=6an-9an-1(n=5≥2,n∈N*),

bn=an+1-ban (n∈N*),若｛cn｝满足c1=1,c2=5,cn+2=5cn+1-6cn (n∈N*),试用数学归纳法证明cn+acn-1=an／(3n-2)(n≥2,n∈N*)

Answer 1

0 = x^2 - x - 6 = (x-3)(x+2),
a = -2 < 3 = b.

a(n+1) = 6a(n) - 9a(n-1),
a(n+2) = 6a(n+1) - 9a(n),
a(n+2) - 3a(n+1) = 3[a(n+1) - 3a(n)],
{a(n+1)-3a(n)}是首项为a(2) - 3a(1) = -6a - 3 = 12 - 3 = 9, 公比为3的等比数列。
a(n+1) - 3a(n) = 9*3^(n-1) = 3^(n+1).

a(n+1) = 3a(n) + 3^(n+1),
a(n+1)/3^(n+1) = a(n)/3^n + 1,
{a(n)/3^n}是首项为a(1)/3 = 1/3, 公差为1的等差数列。
a(n)/3^n = 1/3 + (n-1) = (3n-2)/3,
a(n) = (3n-2)3^(n-1).

b(n) = a(n+1) - ba(n) = a(n+1) - 3a(n) = 3^(n+1).

c(n+2) = 5c(n+1) - 6c(n),
c(n+2) - 2c(n+1)= 3[c(n+1) - 2c(n)],
{c(n+1) - 2c(n)}为首项为c(2) - 2c(1) = 3,公比为3的等比数列。
c(n+1) - 2c(n) = 3*3^(n-1) = 3^n

c(n) + ac(n-1) = c(n) - 2c(n-1) = 3^(n-1) = (3n-2)3^(n-1)/(3n-2) = a(n)/(3n-2)...

用归纳法证明略。。