I tried subtracting $1^{2}+3^{2}….n^{2}$ from $2^{2}+4^{2}….2n^{2}$, but that didn't work.
I know the answer is 
2026-03-26 19:32:19.1774553539
Find $2^{2}+4^{2}….(2n)^{2}$?
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Hint: $\displaystyle 2^2 + 4^2 + \cdots + (2n)^2 = 2^2 (1^2 + 2^2 + \cdots + n^2)$. Plug in the value for $\displaystyle \sum_{k=1}^{n} k^2$ and you'll have the answer.