Use the recursive definition, $f(k) + k² + 2k -3$, to prove that, for any $n ∈ \Bbb N,\ f(n) = n² - 4$.

Question

Here's what I have so far:

Let $f: \mathbb{N} - \{1\} \to \mathbb{N}$, such that $f(x) = (x+2)(x-2)$.
Formulate a recursive definition for $f$ including both the base case $f(2) = 0$ and a $(k+1)^{th}$ case.

$f(2) = 0 $

$\begin{aligned}f(k+1) &= f(k) + ((k+1)+2)((k+1)-2)\\ &= f(k) + (k+3)(k-1)\\ &= f(k) + k^2 + 2k - 3\end{aligned}$

Use the recursive definition from above to prove that, for any $n ∈ \mathbb{N}$, $f(n) = n^2 - 4$.

Base case:
$f(2) = 0 \Rightarrow (4)(0) \Rightarrow (2+2)(2-2)$
so the base case holds.

Induction step:
Assume that for some $k \in \mathbb{N}$, $f(k) = (k+2)(k-2)$
$f(k+1) = f(k) + k^2 + 2k - 3$
$f(k+1) = (k+2)(k-2) + k^2 + 2k - 3$
$f(k+1) = k^2 - 4 + k^2 + 2k - 3 $ $f(k+1) = 2k^2 + 2k - 7$

Any help would be much appreciated.

Answer 1

Here's what I would do: $$ f(k) = k^2 - 4 \implies f(k+1) = (k+1)^2 - 4 = k^2 + 2k - 3, $$ so $f(k+1)=f(k) + 2k + 1$.