Sum of Squares Calculation: 1² + ... + 21²

This problem asks us to find the sum of the squares of the first 21 natural numbers. The series is:

$$ 1^2 + 2^2 + 3^2 + \dots + 21^2 $$

We can calculate this sum using a well-known formula for the sum of the first n squares.

Formula for Sum of Squares

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

$$ \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} $$

In this problem, we need to find the sum up to $21^2$, so we have n = 21.

Applying the Formula

Let's substitute n = 21 into the formula:

1. Calculate n + 1:

   If $n = 21$, then $n + 1 = 21 + 1 = 22$.

2. Calculate 2n + 1:

   If $n = 21$, then $2n + 1 = 2 \times 21 + 1 = 42 + 1 = 43$.

3. Substitute values into the formula:

   Sum = $ \frac{n(n+1)(2n+1)}{6} = \frac{21 \times 22 \times 43}{6} $

4. Simplify the expression:

   We can simplify the calculation by dividing the terms in the numerator by 6.

   Notice that 6 = 2 × 3.

   Divide 21 by 3: $ \frac{21}{3} = 7 $

   Divide 22 by 2: $ \frac{22}{2} = 11 $

   So, the expression becomes: Sum = $ 7 \times 11 \times 43 $

5. Perform the multiplication:

   First, multiply 7 by 11: $ 7 \times 11 = 77 $

   Now, multiply 77 by 43:

   $$ 77 \times 43 $$

   $$ 77 \times 40 = 3080 $$

   $$ 77 \times 3 = 231 $$

   $$ \text{Sum} = 3080 + 231 = 3311 $$

Conclusion

Therefore, the value of the sum $1^2 + 2^2 + 3^2 + \dots + 21^2$ is 3311.