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Question Number 24548 by Physics lover last updated on 20/Nov/17

prove that Σ_(n=1) ^r {n(n−(r/2))^2 }= r∙Σ_(n=1) ^(r/2) n^2 where r = 2k ; k ∈ N

$${prove}\:{that}\: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{{r}} {\sum}}\left\{{n}\left({n}−\frac{{r}}{\mathrm{2}}\right)^{\mathrm{2}} \right\}=\:{r}\centerdot\underset{{n}=\mathrm{1}} {\overset{{r}/\mathrm{2}} {\sum}}{n}^{\mathrm{2}} \\ $$$$\:{where}\:\:\:{r}\:=\:\mathrm{2}{k}\:;\:{k}\:\in\:\mathbb{N} \\ $$

Answered by jota last updated on 21/Nov/17

Σ_1 ^(2k) n(n−k)^2 = Σ(n^3 −2kn^2 +k^2 )=(((2k)^2 (2k+1)^2 )/4) −2k((2k(2k+1)(4k+1))/6)+k^2 ((2k(2k+1))/2)= =2k((k(k+1)(2k+1))/6)=rΣ_1 ^(r/2=k) n^2

$$\: \\ $$$$\:\underset{\mathrm{1}} {\overset{\mathrm{2}{k}} {\sum}}{n}\left({n}−{k}\right)^{\mathrm{2}} = \\ $$$$\:\Sigma\left({n}^{\mathrm{3}} −\mathrm{2}{kn}^{\mathrm{2}} +{k}^{\mathrm{2}} \right)=\frac{\left(\mathrm{2}{k}\right)^{\mathrm{2}} \left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:−\mathrm{2}{k}\frac{\mathrm{2}{k}\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{4}{k}+\mathrm{1}\right)}{\mathrm{6}}+{k}^{\mathrm{2}} \frac{\mathrm{2}{k}\left(\mathrm{2}{k}+\mathrm{1}\right)}{\mathrm{2}}= \\ $$$$\:=\mathrm{2}{k}\frac{{k}\left({k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)}{\mathrm{6}}={r}\underset{\mathrm{1}} {\overset{{r}/\mathrm{2}={k}} {\sum}}{n}^{\mathrm{2}} \\ $$$$ \\ $$

Commented by Physics lover last updated on 21/Nov/17

thank you so much

$${thank}\:{you}\:{so}\:{much} \\ $$

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