Recursive Defintions

$a_{n} = n^{2}$ $a_{0} = 0$ $a_{1} = 1$ $a_{2} = 4$ $base case : a_{0} = 0$ $a_{n} = a_{n - 1} + (2 n - 1)$

Fibonacci Proof (Strong Induction)

Prove $f_{1}^{2} + f_{2}^{2} + ... + f_{n}^{2} = f_{n} f_{n + 1}$

base case: n = 1 Fibonacci numbers: 0 1 1 2 3 5 etc

$0^{2} + 1^{2} = 1 * 1$ ~ the base case holds true

Induction Step:

$f_{1}^{2} + f_{2}^{2} + ... + f_{n}^{2} + f_{n + 1}^{2} = f_{n} f_{n + 1} + f_{n + 1}^{2}$

factor out $f_{n + 1}$ on right side:

$f_{n + 1} (f_{n} + f_{n + 1})$

Notice the expression in the parenthesis:

$f_{n + 1} f_{n + 2}$

The proof is complete

this note was imported from my other vault, ideas are not complete :(

(¬_¬")