Sum of the squares of first n natural numbers exceeds their sum by 330, then n =

The sum of the squares of the first n natural numbers is given by the formula n(n+1)(2n+1)/6.

The sum of the first n natural numbers is given by the formula n(n+1)/2.

According to the problem, the sum of the squares of the first n natural numbers exceeds their sum by 330. So, we can set up the following equation:

n(n+1)(2n+1)/6 - n(n+1)/2 = 330

Solving this equation will give us the value of n.

First, simplify the equation:

n(n+1)[(2n+1)/6 - 1/2] = 330

This simplifies to:

n(n+1)(n-2)/6 = 330

Multiply both sides by 6 to get rid of the fraction:

n(n+1)(n-2) = 1980

This is a cubic equation, and solving it is a bit tricky. But we can guess and check to find that n=10 is a solution.

So, n = 10.