Struggling with a strong induction problem involving recurrence relations The problem states:

Let     $b_0$  = $12$ and  $b_1$ = $29$,  and for all integers  $k ≥ 2$, let  $b_k$ = $5b_{k-1}-6b_{k-2}$.
Prove that for all $n ≥ 0$, $b_n$ = $5\:\cdot \:3^n\:+\:7\:\cdot\:2^n$

I'm familiar with questions where I'm supposed to find the close form of a recurrence relation, and then prove that formula via induction. This question is more strangely structured, however, and I'm not sure how to proceed. Am I supposed to show that:

$b_k$ = $5b_{k-1}-6b_{k-2}$ = $b_n$ = $5\:\cdot \:3^n\:+\:7\:\cdot
\:2^n$

i.e. use strong induction to show both statements are equivalent?
If that's the case, do I show $b_n$ = $5\:\cdot \:3^n\:+\:7\:\cdot \:2^n$ holds for n = 0 and 1 (establishing base cases) and then manipulate what emerges algebraically to equal $b_k$ = $5b_{k-1}-6b_{k-2}$ ? Or should I do that to $b_k$ = $5b_{k-1}-6b_{k-2}$? Advice and insight is most welcome!
 A: First, show that this is true for $n=0$ and $n=1$:
$b_{0}=5\cdot3^{0}+7\cdot2^{0}$
$b_{1}=5\cdot3^{1}+7\cdot2^{1}$
Second, assume that this is true for $n-2$ and $n-1$:
$b_{n-2}=5\cdot3^{n-2}+7\cdot2^{n-2}$
$b_{n-1}=5\cdot3^{n-1}+7\cdot2^{n-1}$
Third, prove that this is true for $n$:
$b_{n}=$
$5\color\red{b_{n-1}}-6\color\green{b_{n-2}}=$
$5(\color\red{5\cdot3^{n-1}+7\cdot2^{n-1}})-6(\color\green{5\cdot3^{n-2}+7\cdot2^{n-2}})=$
$25\cdot3^{n-1}+35\cdot2^{n-1}-30\cdot3^{n-2}-42\cdot2^{n-2}=$
$25\cdot3^{n-1}+35\cdot2^{n-1}-10\cdot3^{n-1}-21\cdot2^{n-1}=$
$15\cdot3^{n-1}+14\cdot2^{n-1}=$
$5\cdot3^{n}+7\cdot2^{n}$

Please note that the assumption is used only in the parts marked red and green.
A: I think you just have to show that $b_n = 5\cdot 3^n + 7\cdot 2^n$ fulfills the recurrence relation and initial conditions for every instance $n$.
As the relation involves up to two prior sequence elements I would use strong induction.
You start with direct proofs for $b_0$, then $b_1$ and the induction from $n=2$.
Update: It seems I need to provide a full solution:
The initial conditions hold by evaluation and comparison.
$$
S(0): b_0 = 12 \wedge 
b_0 = 5\cdot 3^0 + 7\cdot 2^0 = 5 + 7 = 12 \\
S(1): b_1 = 29 \wedge 
b_1 = 5\cdot 3^1 + 7\cdot 2^1 = 15 + 14 = 29 \\
$$
Where the $\wedge$ means logical "and".
The statement for $n$, $n\ge 2$ is
$$
S(n): b_n = 5 b_{n-1} - 6 b_{n-2} \wedge 
b_n = 5\cdot 3^n + 7\cdot 2^n
$$
Base case $n=2$:
$$
S(2):b_2  = 5 b_1 - 6 b_0 \wedge b_2 = 5 \cdot 3^2 + 7 \cdot 2^2
$$
This evaluates to
$$
S(2): b_2 = 5 \cdot 29 - 6 \cdot 12 = 145-72=73 \wedge b_2 = 45 + 28 = 73
$$
which is a true statement.
Induction Step:
Assuming $S(k)$ is true for $k\in\{2,\dotsc, n\}, n\ge 2$ we have
$$
S(k+1): b_{k+1} = 5 b_k - 6 b_{k-1} \wedge 
b_{k+1} = 5\cdot 3^{k+1} + 7\cdot 2^{k+1}
$$
where
\begin{align}
b_{k+1} 
&= 5 b_k - 6 b_{k-1} \\
&= 5 \left( 5\cdot 3^k + 7\cdot 2^k \right) - 
6 \left( 5\cdot 3^{k-1} + 7\cdot 2^{k-1} \right)
\end{align}
The first term in parentheses is justified by the truth of $S(k)$.
The second term needs a case distinction:
For $k=2$ it relies on the truth of the fulfillment of the base condition $b_1$, otherwise we can build on the true statements $S(k-1)$.
Then:
$$
\begin{align}
b_{k+1} 
&= (25 - 10) 3^k + (35-21) 2^k \\
&= 15\cdot 3^k + 14\cdot 2^k \\
&= 5\cdot 3^{k+1} + 7\cdot 2^{k+1}
\end{align}
$$
so we end up with
$$
S(k+1): b_{k+1} = 5\cdot 3^{k+1} + 7\cdot 2^{k+1} \wedge
b_{k+1} = 5\cdot 3^{k+1} + 7\cdot 2^{k+1}
$$
which is a true statement. 
By the principle of strong induction we conclude the truth of $S(n)$ for all integer $n$ with $n>=2$.
Adding the truth of the initial conditions we can claim the truth of
$S(n)$ for all integer $n$ with $n \ge 0$.
Now
\begin{align}
S(0)&: b_0 = 12 \wedge 
b_0 = 5\cdot 3^0 + 7\cdot 2^0 \\
S(1)&: b_1 = 29 \wedge 
b_1 = 5\cdot 3^1 + 7\cdot 2^1 \\
S(n)&: b_n = 5 b_{n-1} - 6 b_{n-2} \wedge 
b_n = 5\cdot 3^n + 7\cdot 2^n \quad (n \ge 2)
\end{align}
means that for all integer $n$ with $n\ge 0$ the sequence $b_n = 5\cdot 3^n + 7\cdot 2^n$ is the solution of the recurrence relation.
