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Question Number 180866 by Mastermind last updated on 18/Nov/22

Answered by mr W last updated on 18/Nov/22

$$1+2+3+...+n=\frac{n(n+1)}{2}$$$$1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$$$${\left[\frac{n(n+1)}{2}\right]}^2=\frac{n(n+1)(2n+1)}{6}+{\psi }_n$$$${\psi }_n=\frac{n^2{(n+1)}^2}{4}-\frac{n(n+1)(2n+1)}{6}$$$${\psi }_n=\frac{3n^4-3n^2+2n^3-2n}{12}$$$${\psi }_n=\frac{n^4-n^2}{4}+\frac{n^3-n}{6}$$$$\Rightarrow \alpha =4,\beta =6$$

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