I want to prove

$\displaystyle (n + 1)^2 + (n + 2)^2 + (n + 3)^2 + \ldots + (2n)^2 = \frac{n(2n + 1)(7n + 1)}{6}, \forall n \geq 1$

or

$\displaystyle \sum_{i=1}^{n} (n + i)^2 = \frac{n(2n + 1)(7n + 1)}{6}, \forall n \geq 1$

by mathematical induction.

I tried with n = 1 and I got

Left side: $\displaystyle (1 + 1)^2 = 2^2 = 4$

Right side: $\displaystyle \frac{1(2 + 1)(7 + 1)}{6} = \frac{3 * 8}{6} = 4$

4 = 4, so I assume $\displaystyle k = n - 1 \implies k^2 + (k + 1)^2 + (k + 2)^2 + \ldots + (2(k-1))^2 = \frac{k(2k + 1)(7k + 1)}{6}, \forall k > 1$

Now I want to prove it for $\displaystyle n = k$ but I don't even know how to start... What should I do? Thanks!