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=((n(n+1))/2) 1^2 +2^2 +3^2 +...+n^2 =((n(n+1)(2n+1))/6) [((n(n+1))/2)]^2 =((n(n+1)(2n+1))/6)+ψ_n ψ_n =((n^2 (n+1)^2 )/4)−((n(n+1)(2n+1))/6) ψ_n =((3n^4 −3n^2 +2n^3 −2n)/(12)) ψ_n =((n^4 −n^2 )/4)+((n^3 −n)/6) ⇒α=4, β=6](https://www.tinkutara.com/question/Q180870.png)
$$\mathrm{1}+\mathrm{2}+\mathrm{3}+…+{n}=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +…+{n}^{\mathrm{2}} =\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$\left[\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right]^{\mathrm{2}} =\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}+\psi_{{n}} \\ $$$$\psi_{{n}} =\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}−\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$\psi_{{n}} =\frac{\mathrm{3}{n}^{\mathrm{4}} −\mathrm{3}{n}^{\mathrm{2}} +\mathrm{2}{n}^{\mathrm{3}} −\mathrm{2}{n}}{\mathrm{12}} \\ $$$$\psi_{{n}} =\frac{{n}^{\mathrm{4}} −{n}^{\mathrm{2}} }{\mathrm{4}}+\frac{{n}^{\mathrm{3}} −{n}}{\mathrm{6}} \\ $$$$\Rightarrow\alpha=\mathrm{4},\:\beta=\mathrm{6} \\ $$