Let $f:\mathbb{N} \rightarrow \mathbb{N}$ be defined by $f(1) = 5, f(2) = 13$, and for $n \ge 2, f(n) = 2f(n - 2) + f(n - 1)$. Prove that $f(n) = 3\cdot 2^n + (-1)^n$ for all $n \in N$
So far I've proved that $f(n)$ is true when $n = 1, 2$. For $k \ge 3$, assume that $p(j)$ is true for all $j \in N, j < k$
Now I want to prove that $p(k)$ is true for all $k \in N$
How would I go about doing that?
Let us do an induction on $n$.
Induction start. $f(1)=5=3\cdot 2-1$ and $f(2)=13=3\cdot 4+1$ (note that we really need the first two values in the induction start, because the recurrence uses the preceeding two values).
Induction hypothesis. For some $n\ge 1$, we assume that we proved the formula for all $k\le n$.
Induction step. We show that the formula then also holds for $n+1\ge 2$:
$$f(n+1)=2f(n-1)+f(n)=2\cdot (3\cdot 2^{n-1}+(-1)^{n-1})+3\cdot 2^n + (-1)^n=3\cdot 2^{n+1}+(-1)^{n+1}$$
Hint $ $ Let $\,S\:\! g(n) = g(n\!+\!1)$ be the $n$-shift operator. $\,\color{#0a0}{(S-2)(2^n)} = 0 = \color{#c00}{(S+1)(-1^n)}\,$ so their product $\,\color{#0a0}{(S-2)}\color{#c00}{(S+1)} = S^2\!-S-2\,$ kills $\, f_n = c\,\color{#0a0}{2^n} + d \color{#c00}{(-1)^n}\,$ for $\,c,d\,$ independent of $\,n.$ Therefore, we deduce $\, 0 = (S^2\!-S-2)f_n = f_{n+2} - f_{n} - 2f_n,\ $ i.e $\ f_{n+2} = f_{n+1} + 2 f_n.$
Remark $ $ See this answer for another example and further explanation (including the above implicit use of commutativity of $\,S-2\,$ and $\,S+1,\,$ which holds since they have constant coef's). This explains the genesis of the common inductive proof in J.R.'s aanswer.
Hint: Try induction. Assume that the statement holds for $n$ and from that, prove it holds for $n+1$.
