Square of Sum with Double/Algebraic Proof 2

Theorem

$\forall a, b \in \R: \paren {a + 2 b}^2 = a^2 + 4 a b + 4 b^2$

Proof

A direct application of the Binomial Theorem:

$\ds \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$

putting $n = 2, x = a, y = 2 b$.

$\blacksquare$