Problem:
Assume that the terms of the sequence $(a_n)^{\infty}_{n=1}$ satisfy $a_1 =6$ and:
$$a_{n+1}=3a_n + 8^n \text{ for every } n \ge 1$$
Use the principle of mathematical induction to prove that the equality $a_n=\frac{22}{15} \cdot 3^n + \frac15 \cdot 8^n$ holds for every positive integer $n$.
What I have so far:
base case: when $n=1$
$a_1=\frac{22}{15} \cdot 3^1 + \frac15 \cdot 8^1$, this equals $6$.
Hypothesis: assume $n=k$ for $n \ge 1$, $a_k=\frac{22}{15} \cdot 3^k + \frac15 \cdot 8^k$
Induction: Prove for $k+1$, $a_{k+1}=\frac{22}{15} \cdot 3^{k+1} + \frac15 \cdot 8^{k+1}$
But now I am confused on what I have to show? Do I have to manipulate $a_{k+1}$ and make it look like $a_{n+1}=3a_n + 8^n$. Or do I substitute $a_k$ for $a_n$ and get $a_{k+1}=3a_k + 8^k$ and make this look like $a_{k+1}$?
So, $3(\frac{22}{15} \cdot 3^k + \frac15 \cdot 8^k) + 8^k=\frac{22}{15} \cdot 3^{k+1} + \frac15 \cdot 8^{k+1}$
Any help on what the next step is would be greatly appreciated.
The idea with induction is to take the proposition for $k$ and prove it for $k+1$, in addition to proving the proposition for $k=1$. Your base case ($k=1$) is correct.
You start with
$$a_k = \frac{22}{15}3^k + \frac{1}{5}8^k$$
Now input this into the recurrence relation:
$$a_{k+1} = 3\left(\frac{22}{15}3^k + \frac{1}{5}8^k\right)+8^k$$
$$a_{k+1} = 3\cdot\frac{22}{15}3^k + 3\cdot\frac{1}{5}8^k+8^k$$
$$a_{k+1} = \frac{22}{15}3^{k+1} + \frac{3}{5}8^k+8^k$$
$$a_{k+1} = \frac{22}{15}3^{k+1} + \frac{8}{5}8^k$$
$$a_{k+1} = \frac{22}{15}3^{k+1} + \frac{1}{5}8^{k+1}$$
thus proving the result for $k+1$ and completing the inductive step.