Question Number 186630 by DAVONG last updated on 07/Feb/23
Answered by mr W last updated on 07/Feb/23
$${a}_{{k}} =\mathrm{1}+\mathrm{2}+\mathrm{3}+...+{k}=\frac{{k}\left({k}+\mathrm{1}\right)}{\mathrm{2}}=\frac{{k}^{\mathrm{2}} +{k}}{\mathrm{2}} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{k}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} +\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\right) \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{12}}+\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{4}} \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{\mathrm{6}} \\ $$$${with}\:{n}=\mathrm{2022}: \\ $$$$\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{\mathrm{6}} \\ $$$$=\frac{\mathrm{2022}×\mathrm{2023}×\mathrm{2024}}{\mathrm{6}}=\mathrm{1}\:\mathrm{379}\:\mathrm{864}\:\mathrm{024} \\ $$