The sum of squares of the first ten natural numbers is:

Correct option is C

Given:

We need to find the sum of squares of the first ten natural numbers.

Formula Used:

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

​$$S = \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$$​​

Solution:

For n=10:

S$$= \frac{10(10+1)(2 \times 10 + 1)}{6}$$​​

S= $$\frac{10 \times 11 \times 21}{6}$$​​

S =$$\frac{2310}{6} = 385$$

Option (C) is right.