Find the value of $$21^2+22^2+23^2+........+30^2$$​​

Correct option is B

Given:

Sum of squares: 21² + 22² + 23² + ... + 30²

Formula Used:

The sum of squares of the first n natural numbers is given by:
​$$\sum_{k=1}^n k^2 = \frac{n(n + 1)(2n + 1)}{6}$$​
Solution:
$$\sum_{k=1}^{30} k^2 = \frac{30 \times 31 \times 61}{6} = \frac{56730}{6} = 9455$$​​

$$\sum_{k=1}^{20} k^2 = \frac{20 \times 21 \times 41}{6} = \frac{17220}{6} = 2870$$

​$$\sum_{k=21}^{30} k^2 = \sum_{k=1}^{30} k^2 - \sum_{k=1}^{20} k^2 = 9455 - 2870 = 6585$$​​