The sum of the first $n$ terms of the series $\frac{1^2}{1} + \frac{2^2}{1+2} + \frac{3^2}{1+2+3} + \ldots$

Question

$\frac{n+2}{3}$

$\frac{n(n+2)}{3}$

$\frac{n(n-2)}{3}$

$\frac{n(n-2)}{6}$

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