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Question Number 206993 by necx122 last updated on 03/May/24

If the nth term of a sequence is given by ((n^2 −2n)/4) ,what is the sum of n terms of the sequence?

$${If}\:{the}\:{nth}\:{term}\:{of}\:{a}\:{sequence}\:{is} \\ $$$${given}\:{by}\:\frac{{n}^{\mathrm{2}} −\mathrm{2}{n}}{\mathrm{4}}\:,{what}\:{is}\:{the}\:{sum}\:{of}\:{n} \\ $$$${terms}\:{of}\:{the}\:{sequence}? \\ $$

Answered by Rasheed.Sindhi last updated on 03/May/24

Σ_(k=1) ^n ((k^2 −2k)/4) =Σ_(k=1) ^n ((k^2 /4)−(k/2)) =(1/4)Σ_(k=1) ^n k^2 −(1/2)Σ_(k=1) ^(n) k =(1/4)(((n(n+1)(2n+1))/6))−(1/2)(((n(n+1))/2)) =((n(n+1)(2n+1))/(24))−((n(n+1))/4) =n(n+1)(((2n+1)/(24))−(1/4)) =n(n+1)(((2n+1−6)/(24))) =(1/(24))n(n+1)(2n−5)

$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{k}^{\mathrm{2}} −\mathrm{2}{k}}{\mathrm{4}} \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{{k}^{\mathrm{2}} }{\mathrm{4}}−\frac{{k}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\Sigma}}{k} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right) \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{24}}−\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{4}} \\ $$$$={n}\left({n}+\mathrm{1}\right)\left(\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{24}}−\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$={n}\left({n}+\mathrm{1}\right)\left(\frac{\mathrm{2}{n}+\mathrm{1}−\mathrm{6}}{\mathrm{24}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{24}}{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{5}\right) \\ $$

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?

$${Thank}\:{you}\:{so}\:{much}\:{sir}.\:{Also},\:{is}\:{there} \\ $$$${a}\:{way}\:{to}\:{prove}\:{what}\:{sum}\:{of}\:{a}\:{square} \\ $$$${function}? \\ $$

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