Question

If the sum of the squares of first n natural numbers exceeds their sum by 330, then n = _______________.

Answer 1

Sum of squares of first n natural numbers is $\frac{n\left(n+1\right)\left(2n+1\right)}{6}$
Sum of first n natural number is $\frac{n\left(n+1\right)}{2}$
According to given condition, $\mathrm{\Sigma }{n}^2-\mathrm{\Sigma }n=330$
$$\begin{gathered} \mathrm{i}.\mathrm{e}\frac{n\left(n+1\right)\left(2n+1\right)}{6}-\frac{n\left(n+1\right)}{2}=330 \\ \mathrm{i}.\mathrm{e}\frac{n\left(n+1\right)}{2}\left[\frac{2n+1}{3}-1\right]=330 \\ \mathrm{i}.\mathrm{e}\frac{n\left(n+1\right)}{2}\left[\frac{2n+1-3}{3}\right]=330 \\ \mathrm{i}.\mathrm{e}\frac{n\left(n+1\right)}{2}\left[\frac{2n-2}{3}\right]=330 \\ \mathrm{i}.\mathrm{e}\frac{n\left(n+1\right)2\left(n-1\right)}{2\times 3}=330 \\ \mathrm{i}.\mathrm{e}n\left(n^2-1\right)=330\times 3 \\ \mathrm{i}.\mathrm{e}{n}^3-n=990 \\ \mathrm{i}.\mathrm{e}{n}^3-n-990=0 \end{gathered}$$
i.e n³ − 10n² + 10n² − n − 990 = 0 − 99n + 99n
i.e n² (n − 10) + 10n² − 100n − 990 + 99n = 0
i.e n² (n − 10) + 10n (n − 10) + 99(n−10) = 0
i.e (n − 10) (n² + 10n + 99) = 0
⇒ n = 10