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.

$$\frac{2k(k+1)(2k+1)}{3}+(2(k+1))^2=\frac{2k(k+1)(2k+1)}{3}+\frac{3(2(k+1))^2}{3}=\\=
\frac{2k(k+1)(2k+1)+3 \cdot 4(k+1)^2}{3}=\frac{2(k+1)(k(2k+1)+6(k+1))}{3}=\\=\frac{2(k+1)(2k^2+k+6k+6)}{3}=\frac{2(k+1)((2k+3)(k+2))}{3}$$
