Respuesta
$\sum_{k=1}^n k^2 = \frac{n(n + 1)(2n + 1)}{6}$.
Explicación
Similarly, $\sum_{k=1}^n k = \frac{n(n + 1)}{2}$ and $\sum_{k=1}^n k^3 = \left[\frac{n(n + 1)}{2}\right]^2$. These identities frequently appear in sequence problems.
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$\sum_{k=1}^n k^2 = \frac{n(n + 1)(2n + 1)}{6}$.
Similarly, $\sum_{k=1}^n k = \frac{n(n + 1)}{2}$ and $\sum_{k=1}^n k^3 = \left[\frac{n(n + 1)}{2}\right]^2$. These identities frequently appear in sequence problems.