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First, we see that equation A.1 is true for $n = 0$ (trivial).

Next, assume that the equation A.1 holds for $n = k$:

$$2 \cdot \sum_{i=0}^k i = k (k + 1)$$ (2)

Finally we show that if the equation A.2 is true, then equation A.1 also holds for $n = k + 1$.
That is, we want to show:

$$2 \cdot \sum_{i=0}^{k+1} i = (k+1)(k+2)$$ (3)

Now:

$$2 \cdot \sum_{i=0}^{k+1} = 2 \cdot \bigg( (k+1) + \sum_{i=0}^k i \bigg)$$ (4)

$$= 2(k+1) + 2 \bigg( \sum_{i=0}^k i \bigg)$$ (5)

$$= 2k + 2 + k(k+1)$$ (6)

$$= 2k + 2 + k^2 + k$$ (7)

$$= k^2 +3k + 2$$ (8)

$$= (k+1)(k+2)$$ (9)

$\therefore$ equation A.3 is true.
$\therefore$ Since equation A.1 is true for $n = 0$, then equation A.1 is also true for $n = 1$.
$\therefore$ equation A.1 is true for all $n \geq 0 \in \mathbb{N}$ by the principle of mathematical induction.