<s>[INST] 'Mathematical Induction' Use mathematical induction to prove that $4^n -3^n + 1 = 7a_{n-1} – 12a_{n-2} + 6$  with $n \ge 3$ with the initial condition $a_1 = 2$ and $a_2 = 8$ . Given that $a_n = 4^n -3^n + 1$. 
I am confused at the inductive step or $(k+1)$ part.    
 [/INST] Since the OP asks for an inductive proof, I'll write in this format.
Base case


*

*$n = 3$


*

*$\text{LHS} = 4^3 - 3^3 + 1 = 64 - 27 + 1 = 38$

*$\text{RHS} = 7a_2 - 12a_1 + 6 = 7(8) - 12(2) + 6 = 56 - 24 + 6 = 38$


*$n = 4$


*

*$\text{LHS} = 4^4 - 3^4 + 1 = 256 - 81 + 1 = 176$

*$\text{RHS} = 7 a_3 - 12 a_2 + 6 = 7(38) - 12(8) + 6 = 266 - 96 + 6 = 176$



Induction hypothesis (redundant)


*

*$n = k-1, a_{k-1} := 4^{k-1} - 3^{k-1} + 1 = 7 a_{k-2} - 12 a_{k-3} + 6$

*$n = k, a_k := 4^k - 3^k + 1 = 7 a_{k-1} - 12 a_{k-2} + 6$


Inductive step
\begin{align*}
n &= k+1 \\
a_{k+1} &:= 4^{k+1} - 3^{k+1} + 1 \\
&= 4 \cdot 4^k - 3 \cdot 3^k + 1 \\
&= 7 (4^k - 3^k + 1) - 3 \cdot 4^k + 4 \cdot 3^k - 6 \\
&= 7 a_k - 12 (4^{k-1} - 3^{k-1} + 1) + 6 \\
&= 7 a_k - 12 a_{k-1} + 6
\end{align*}
Hence, by induction, the proposition is true for all $n \in \Bbb N$.
Remarks
The induction hypothesis is not needed.  As vrugtehagel says, what we're doing is just verifying $4^n-3^n+1=7a_{n-1}-12a_{n-2}+6 \,\forall\,n\ge3$.
</s>