We know that (1² + 2² + 3² + ... + $ n $²) =$ \dfrac{1}{6} n $$\left( n + 1\right)$$\left(2 n + 1\right)$

Putting $ n $ = 10, required sum =$ \left(\dfrac{1}{6} \times 10 \times 11 \times 21\right) $= 385