Proving by strong induction for a sequence of integers, $2^n$ divides term $n$ Provided the following sequence of integers $t_1, t_2, t_3$,... is defined as:  
$t_1 =4, t_2 =8$ and $t_n= $ $ 6t_n$$_-$$_1$ - $4t_n$$_-$$_2$ for all integers $n \geq 3$ 
How do we prove that $2^n$ divides $t_n$ for all integers $n \geq 1$ by strong induction?
So far, I am having problems with the induction step as I am still trying to grasp the concept of strong induction, but this is what I came up with so far:
Base step:
$t_3 = 32$, which is divisible by 8; and $t_4 = 160$, which is divisible by 16.
$\therefore$ Statement is true when $n=3$ and $n=4$.
Induction step:
Assuming $k=n$ and the statement is true for $3,4,...k$. Prove that $2^k$ divides $t_k$ for $k+1$:
$t_3 = ((2^3 - 2)(2^3) - (2^2)(2^2)) = 2^6 - 2^4 - 2^4 $; dividing this by $2^3$ gives $2^3 - 2^2$ which would be an integer result.
$t_4 = ((2^3 - 2)(2^5) - (2^2)(2^3)) = 2^8 - 2^6 - 2^5 $; dividing this by $2^4$ gives $2^4 - 2^2 - 2$ which would also be an integer result.
$\therefore$ By the logic above, the statement also holds for $k-1$, $k$ as well as $k+1$, thus as required to prove by strong induction.
I believe this is a fundamentally flawed way of trying to prove by strong induction, but this is the only way I can possibly think of. Can someone show a correct method of solving this?

Let
$u_n = t_n/2^n$,
or
$t_n = 2^n u_n$.
Since
$t_n
=6 t_{n-1}-4t_{n-2}
$,
$2^nu_n
=6 \cdot 2^{n-1}u_{n-1}-4\cdot 2^{n-2}u_{n-2}
=3 \cdot 2^{n}u_{n-1}- 2^{n}u_{n-2}
$,
or
$u_n
=3 u_{n-1}- u_{n-2}
$.
Since $u_1$ and $u_2$
are integers
(
$u_1 = t_1/2 = 2$
and
$u_2 = t_2/4 = 2$
),
all following
$u_n$ are integers.
Even more,
since $u_1$ and $u_2$
are even integers,
so are all following
$u_n$.
