So approaching this problem
For all integers $n ≥ 0$, $y_n = 3 * 2^n + 4^n$ where $y_0 = 4, y_1 = 10$ and when $n ≥ 2$, $y_n = 6y_{n-1} – 8y_{n-2} $
I realise that is probobly needs to be proved by induction I have the base cases done for $y_0=4$ and $y_1=10$ but I am having a hard time finding out the inductive step I need to take.
Assume this holds for all $y_0, y_1, \ldots y_k$ with $k \ge 1$ and let's prove it for $k \ge 2$. We have $$ \begin{split} y_k &= 6y_{k-1} - 8y_{k-2} \\ &= 6\left(3 \cdot 2^{k-1} + 4^{k-1}\right) - 8\left(3 \cdot 2^{k-2} + 4^{k-2}\right) \\ &= 9\cdot 2^k + \frac{3}{2} 4^k - 6\cdot 2^k - \frac{1}{2} 4^k \\ &= 3\cdot 2^k + 4^k, \end{split} $$ as desired.
For the sake of completeness, this is a recurrence relation with constant coefficients, so you could solve it analytically without having to guess and use induction. So let $y_k - 6y_{k-1} + 8y_{k-2} = 0$ with the initial conditions $y_0=4$ and $y_1=10$.
Assuming $y_k = a^k$ we get the characteristic equation $$ 0 = a^2-6a+8 = (a-2)(a-4) $$ so the roots are $a=2$ and $a=4$, which implies $$ y_n = A\cdot 2^n + B \cdot 4^n $$ and now apply the initial conditions to find a linear system of 2 unknowns and 2 equations for $A$ and $B$: $$ \begin{split} 4 &= y_0 &= A &+ B\\ 10 &= y_1 &= 2A &+ 4B \end{split} $$ The second equation is equivalent to $A+2B=5$ and subtracting the first one yields $B=1$, plugging back into the the first one yields $A=4-1=3$. Hence, $$ y_n = 3\cdot 2^n + 4^n $$