Find the sum of the numbers

Relevant wiki: Sum of n, n², or n³

The sum of the squares of the first n natural numbers is given by the formula,

1² + 2² + 3² + 4² ....... + n² = $\frac{n(n+1)(2n + 1)}{6}$

So, n = 40 ; [ 40² = 1600 ]

The sum is,

= $\frac{40 × 41 × 81}{6}$

= 20 × 27 × 41

= 22140 (Answer)

Sum off $\sum \limits_{n=1}^{n} n^2 = \dfrac{(n)(n+1)(2n+1)}{6}$ . We observe that this series is also a series involving sum of square till $n=40$ . Substitute in formula, we get

$\dfrac{(40)(41)(81)}{6} = \boxed{22140}$