Use $\sum_{k=1}^{n} k^{2} = \dfrac{n(n+1)(2n+1)}{6}$ to find $1^{2} + 2^{2} + \ldots + 10^{2}$.

Use $\sum_{k=1}^{n} k^{2} = \dfrac{n(n+1)(2n+1)}{6}$ to find $1^{2} + 2^{2} + \ldots + 10^{2}$.
  1. $385$
  2. $360$
  3. $405$
  4. $350$

Solution

$\dfrac{10 \cdot 11 \cdot 21}{6} = \dfrac{2310}{6} = 385$.

Asked in: IMO

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