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Question Number 120480 by mathocean1 last updated on 31/Oct/20
show that ∀ n ∈N^∗ Σ_(k=1) ^n k(n−k)=(((n−1)(n+1))/6)
$${show}\:{that}\:\forall\:{n}\:\in\mathbb{N}^{\ast} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} {k}\left({n}−{k}\right)=\frac{\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$
Answered by Ar Brandon last updated on 31/Oct/20
S_k =Σ_(k=1) ^n k(n−k)=Σ_(k=1) ^n (kn−k^2 ) =((n^2 (n+1))/2)−((n(n+1)(2n+1))/6) =((3n^3 +3n^2 −(2n^3 +3n^2 +n))/6) =((n^3 −n)/6)=((n(n−1)(n+1))/6)
$$\mathrm{S}_{\mathrm{k}} =\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{k}\left(\mathrm{n}−\mathrm{k}\right)=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\mathrm{kn}−\mathrm{k}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:=\frac{\mathrm{n}^{\mathrm{2}} \left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}}−\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{3n}^{\mathrm{3}} +\mathrm{3n}^{\mathrm{2}} −\left(\mathrm{2n}^{\mathrm{3}} +\mathrm{3n}^{\mathrm{2}} +\mathrm{n}\right)}{\mathrm{6}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{n}^{\mathrm{3}} −\mathrm{n}}{\mathrm{6}}=\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$
Answered by Dwaipayan Shikari last updated on 31/Oct/20
Σ_(k=1) ^n kn−Σ_(k=1) ^n k^2 =((n^2 (n+1))/2)−((n(n+1)(2n+1))/6) =((n(n+1))/6)(3n−2n−1)=((n(n^2 −1))/6)
$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{kn}−\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} \\ $$$$=\frac{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)}{\mathrm{2}}−\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{6}}\left(\mathrm{3}{n}−\mathrm{2}{n}−\mathrm{1}\right)=\frac{{n}\left({n}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{6}} \\ $$
Commented by mathocean1 last updated on 31/Oct/20
thanks sirs is it also possible to show it by recurrence ?
$${thanks}\:{sirs} \\ $$$$ \\ $$$${is}\:{it}\:{also}\:{possible}\:{to}\:{show}\:{it}\:{by}\:{recurrence}\:? \\ $$
Answered by Ar Brandon last updated on 31/Oct/20
Let P_n be the statement Σ_(k=1) ^n k(n−k)=((n(n−1)(n+1))/6) For n=1, 0=0 (true) For n=2, (2−1)+2(2−2)=1=((2(1)(3))/6)=1 (true) Suppose P_n true for all n∈N^∗ and show that P_(n+1) is true. Σ_(k=1) ^n k(n−k)=((n(n−1)(n+1))/6) P_(n+1) :Σ_(k=1) ^(n+1) k(n−k)=Σ_(k=1) ^n k(n−k)−(n+1) =((n(n−1)(n+1))/6)−(n+1)=(((n+1)(n^2 −n−6))/6) =(((n+1)(n−3)(n+2))/6)
$$\mathrm{Let}\:\mathrm{P}_{\mathrm{n}} \:\mathrm{be}\:\mathrm{the}\:\mathrm{statement} \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{k}\left(\mathrm{n}−\mathrm{k}\right)=\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$\mathrm{For}\:\mathrm{n}=\mathrm{1},\:\mathrm{0}=\mathrm{0}\:\left(\mathrm{true}\right) \\ $$$$\mathrm{For}\:\mathrm{n}=\mathrm{2},\:\left(\mathrm{2}−\mathrm{1}\right)+\mathrm{2}\left(\mathrm{2}−\mathrm{2}\right)=\mathrm{1}=\frac{\mathrm{2}\left(\mathrm{1}\right)\left(\mathrm{3}\right)}{\mathrm{6}}=\mathrm{1}\:\left(\mathrm{true}\right) \\ $$$$\mathrm{Suppose}\:\mathrm{P}_{\mathrm{n}} \:\mathrm{true}\:\mathrm{for}\:\mathrm{all}\:\mathrm{n}\in\mathbb{N}^{\ast} \:\mathrm{and}\:\mathrm{show}\:\mathrm{that}\:\mathrm{P}_{\mathrm{n}+\mathrm{1}} \:\mathrm{is}\:\mathrm{true}. \\ $$$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{k}\left(\mathrm{n}−\mathrm{k}\right)=\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$\mathrm{P}_{\mathrm{n}+\mathrm{1}} :\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}+\mathrm{1}} {\sum}}\mathrm{k}\left(\mathrm{n}−\mathrm{k}\right)=\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{k}\left(\mathrm{n}−\mathrm{k}\right)−\left(\mathrm{n}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{6}}−\left(\mathrm{n}+\mathrm{1}\right)=\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}^{\mathrm{2}} −\mathrm{n}−\mathrm{6}\right)}{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}−\mathrm{3}\right)\left(\mathrm{n}+\mathrm{2}\right)}{\mathrm{6}} \\ $$
Commented by Ar Brandon last updated on 31/Oct/20
I'm not getting it right.
Answered by mathmax by abdo last updated on 31/Oct/20
Σ_(k=1) ^n k(n−k) =nΣ_(k=1) ^n k−Σ_(k=1) ^n k^2 =n((n(n+1))/2)−((n(n+1)(2n+1))/6) =((n(n+1))/2){n−((2n+1)/3)} =((n(n+1))/2)(((3n−2n−1)/3)) =((n(n+1)(n−1))/6)
$$\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \mathrm{k}\left(\mathrm{n}−\mathrm{k}\right)\:=\mathrm{n}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \mathrm{k}−\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \mathrm{k}^{\mathrm{2}} \\ $$$$=\mathrm{n}\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}}−\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)}{\mathrm{6}}\:=\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}}\left\{\mathrm{n}−\frac{\mathrm{2n}+\mathrm{1}}{\mathrm{3}}\right\} \\ $$$$=\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}}\left(\frac{\mathrm{3n}−\mathrm{2n}−\mathrm{1}}{\mathrm{3}}\right)\:=\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}−\mathrm{1}\right)}{\mathrm{6}} \\ $$

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