Edit: Apparently my question wasn't clear enough so I will go step by step:
I will start from the beginning:
Claim : Let $a_n$ be the sequence defined by $a_1=1, a_2=8,$ and $a_n= a_{n-1} + 2a_{n-2} $ , for $ n \ge 3$
Proof: Prove by strong induction that, for all $ n \in N$,
(*) $a_n= 3 \cdot2^{n-1} + 2(-1)^n$
Base case: When $n= 1$, the left side is $a_1=1$, and the right side is $3 \cdot2^0 + 2(-1)^1 =1 $ . So both sides are equal and (*) is true for $n=1$.
Same is the case with $n=2$. Now I move to induction step.
Induction step: Let $k \in N $ with $k \ge 2 $ be given and suppose (*) is true for n = 1, 2 ..... k. Then:
$a_{k+1} = a_k + 2a_{k-1}$ (by recurrence of $a_n$ )
$=3*2^{k-1} + 2(-1)^k + 2(3*2^{k-2} + 2(-1)^{k-1}) $ (by $n=k$ and $n= k-1$)
This is as far as I got in solving the problem. My question is how do I get to final solution: $3*2^k + 2(-1)^{k+1}$ from the last step I made.
I tried myself but I have gaps in my Algebra knowledge, so I always got different outcome. In other words, my question is can someone solve a problem above from where I left it, step by step, pointing out why and what he used to get to the solution.
P.S I hope this is clear enough now.If not, please let me know what you didn't understand like @ctst did. Also this is my first time, I ever asked online for a help to solve a math problems or used Math Stackexchange.
Yes it is clear enaugh, indeed. I don't think you still deserve downvotes.
So, you found that $$a_{k+1}=3\times 2^{k-1}+2(-1)^k+2\times \left(3\times 2^{k-2}+2(-1)^{k-1}\right)\tag{1}$$ Your final result should be $$3\times2^k+2(-1)^{k+1}\tag{2}$$ Notice that this final expression is a sum of two terms: $3$ times a power of $2$ and $2$ times a power of $(-1)$. So why not try to regroup powers of $2$ and powers of $(-1)$ in expression $(1)$, like this:$$a_{k+1}=\left(3\times 2^{k-1}+2\times 3\times 2^{k-2}\right)+\left( 2(-1)^k+2\times 2(-1)^{k-1}\right)$$
All you need now is to simplify this expression. You can do it!