Question Number 206993 by necx122 last updated on 03/May/24
$${If}\:{the}\:{nth}\:{term}\:{of}\:{a}\:{sequence}\:{is} \\ $$$${given}\:{by}\:\frac{{n}^{\mathrm{2}} −\mathrm{2}{n}}{\mathrm{4}}\:,{what}\:{is}\:{the}\:{sum}\:{of}\:{n} \\ $$$${terms}\:{of}\:{the}\:{sequence}? \\ $$
Answered by Rasheed.Sindhi last updated on 03/May/24
$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{k}^{\mathrm{2}} −\mathrm{2}{k}}{\mathrm{4}} \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{{k}^{\mathrm{2}} }{\mathrm{4}}−\frac{{k}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\Sigma}}{k} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right) \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{24}}−\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{4}} \\ $$$$={n}\left({n}+\mathrm{1}\right)\left(\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{24}}−\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$={n}\left({n}+\mathrm{1}\right)\left(\frac{\mathrm{2}{n}+\mathrm{1}−\mathrm{6}}{\mathrm{24}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{24}}{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{5}\right) \\ $$
Commented by necx122 last updated on 04/May/24
$${Thank}\:{you}\:{so}\:{much}\:{sir}.\:{Also},\:{is}\:{there} \\ $$$${a}\:{way}\:{to}\:{prove}\:{what}\:{sum}\:{of}\:{a}\:{square} \\ $$$${function}? \\ $$