mathematical induction (sum of squares of first $2n$ numbers)

I need to prove with mathematical induction the formula for the sum of the squares of the first $2n$ numbers.
The equation:
$$1^2 + 2^2 + 3^2 + \cdots + (2n)^2 = \dfrac{n(2n+1)(4n+1)}{3}.$$

The equation stands for $n \geqslant 1$.

Here is my solution:

$n=1$:
$1^2 + 2^2 = \dfrac{1 \cdot 3 \cdot 5}{3}$
$5 = 5$

$n = k$:
$1^2+2^2+\cdots+(2k)^2 = \dfrac{k(2k+1)(4k+1)}{3}$

$n = k + 1$:
$1^2+2^2+\cdots+(2k)^2 +(2к + 1)^2 + (2к + 2)^2 = \dfrac{(k+1)(2(k+1)+1)(4(k+1)+1)}{3}$

$\dfrac{k(2k+1)(4k+1)}{3} + \dfrac{3(2k + 1)^2}{3} + \dfrac{3(2k + 2)^2}{3} = \dfrac{(k+1)(2(k+1)+1)(4(k+1)+1)}{3}$

$\dfrac{8k^3 + 2k^2 + 4k^2 + k}{3} + \dfrac{3(4k^2 + 4k + 1)}{3} + \dfrac{3(4k^2 + 8k + 4)}{3} = \dfrac{(k+1)(2k+3)(4k+5)}{3}$

$\dfrac{8k^3 + 6k^2 + k + 12k^2 + 12k + 3 + 12k^2 + 24k + 12}{3} = \dfrac{8k^3 + 10k^2 + 12k^2 + 15k + 8k^2 + 10k + 12k + 15}{3}$

$\dfrac{8k^3 + 30k^2 + 37k + 15}{3} = \dfrac{8k^3 + 30k^2 + 37k + 15}{3}$

In the end, the terms are not equal, and they should be. What am I doing wrong?

Answer

You are forgetting to add the term $(2k+1)^{2}$ in the LHS,i.e. for $n=k+1$ the LHS is $$1^{2}+2^{2}+\cdots +(2k)^{2}+(2k+1)^{2}+(2k+2)^{2}$$