The sum of the squares of the first n natural | vedclass

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(B) The sum of the squares of the first $n$ natural numbers is given by $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$.The sum of the first $n$ natural numbers i

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$10$

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(B) The sum of the squares of the first $n$ natural numbers is given by $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$.The sum of the first $n$ natural numbers is given by $\sum n = \frac{n(n+1)}{2}$.According to the problem,$\sum n^2 = \sum n + 330$.Substituting the formulas: $\frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)}{2} + 330$.Rearranging the terms: $\frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} = 330$.Taking $\frac{n(n+1)}{2}$ as a common factor: $\frac{n(n+1)}{2} \left[ \frac{2n+1}{3} - 1 \right] = 330$.Simplifying the expression inside the bracket: $\frac{n(n+1)}{2} \left[ \frac{2n+1-3}{3} \right] = 330$.$\frac{n(n+1)}{2} \cdot \frac{2(n-1)}{3} = 330$.$\frac{n(n+1)(n-1)}{3} = 330$.$n(n^2-1) = 990$.$n^3 - n - 990 = 0$.Testing values,if $n=10$,then $10^3 - 10 = 1000 - 10 = 990$. Thus,$n = 10$.

The sum of the squares of the first $n$ natural numbers exceeds their sum by $330$. Then,$n = $

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