<s>[INST] Mathematical Induction: Sum of first n even perfect squares So the series is $$P_k: 2^2 + 4^2 + 6^2 + ... + (2k)^2 = \frac{2k(k+1)(2k+1)}3$$
and i have to replace $P_k$ with $P_{k+1}$ to prove the series.
I have to show that $$\frac{2k(k+1)(2k+1)}3 + [2(k+1)]^2 = \frac{2(k+1)(k+2)(2k+3)}3$$ but I don't know how.
 [/INST] There isn't a simple trick to solve this; it requires a battle with the algebra monster. When doing a proof by induction you have to start with the base case though, so let's do that first. We need to show that $$\sum_{i=1}^n(2i)^2=\frac{2n(n+1)(2n+1)}{3}$$
Base case n=1:
$$\sum_{i=1}^1(2i)^2=4 = \frac{12}{3} = \frac{2 \cdot 2 \cdot 3}{3} = \frac{2\cdot 1(1+1)(2\cdot 1+1)}{3} \quad \checkmark$$
Induction Step: 
Suppose for all $k$ where $1 < k\leq n$ that $$\sum_{i=1}^k(2i)^2=\frac{2n(n+1)(2n+1)}{3}$$
Then we know $$\sum_{i=1}^{k+1}(2i)^2=\sum_{i=1}^{k}(2i)^2+ [2(k+1)]^2$$ Now we can apply the induction hypothesis to the above equality to make the step 
$$\sum_{i=1}^{k}(2i)^2+ [2(k+1)]^2 = \frac{2k(k+1)(2k+1)}3 + [2(k+1)]^2 \\= \frac{2k(k+1)(2k+1)}3+ \frac{3[2(k+1)]^2}{3}  \\ = \frac{2k(k+1)(2k+1)+3[2(k+1)]^2}3 \\ = \frac{2k(k+1)(2k+1)+12(k+1)^2}3$$ Factor out the quantity $2(k+1)$. $$= \frac{2(k+1)[k(2k+1)+6(k+1)]}{3} \\= \frac{2(k+1)(2k^2+k+6k+6)}{3} \\ = \frac{2(k+1)(2k^2+7k+6)}{3} \\= \frac{2(k+1)(k+2)(2k+3)}{3}\quad \quad \Box$$ 
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