Sum of Squares Calculation

This problem requires calculating the sum of squares for a specific range of numbers (3 to 18). We can use the standard formula for the sum of the squares of the first n natural numbers and adapt it for the given range.

Sum of Squares Formula

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

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

Applying the Formula

To find the sum of squares from 3 to 18 ($\sum_{n=3}^{18} n^2$), we first calculate the sum of squares from 1 to 18 and then subtract the sum of squares from 1 to 2 ($\sum_{n=1}^{2} n^2$).

Step 1: Calculate Sum of Squares from 1 to 18

Using the formula with n = 18:

\sum_{i=1}^{18} i^2 = \frac{18(18+1)(2 \times 18 + 1)}{6}

= \frac{18 \times 19 \times (36 + 1)}{6}

= \frac{18 \times 19 \times 37}{6}

= 3 \times 19 \times 37

= 57 \times 37

= 2109

Step 2: Calculate Sum of Squares from 1 to 2

Using the formula with n = 2:

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

= \frac{2 \times 3 \times (4 + 1)}{6}

= \frac{2 \times 3 \times 5}{6}

= \frac{30}{6}

= 5

Step 3: Calculate the Final Sum

Subtract the sum of squares from 1 to 2 from the sum of squares from 1 to 18:

\sum_{n=3}^{18} n^2 = \sum_{i=1}^{18} i^2 - \sum_{i=1}^{2} i^2

= 2109 - 5

= 2104

Conclusion

The sum of the squares of the numbers from 3 to 18 is 2104.