Question

What is the recursive formula for the sequence 4, 16, 36, 64, 100...?

Answer 1

Unrecursively, that's

$f(n) = (2n)^2 = 4n^2$

A quadratic will have a linear first difference:

$f(n+1) - f(n) = 4(n+1)^2 - 4n^2 = 4n^2 + 8n + 4 - 4n^2 = 8n+ 4$

So we get

f(1) = 4

f(n+1) = f(n) + 8n + 4