Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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)

$$\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \mathrm{k}\left(\mathrm{n}−\mathrm{k}\right)\:=\mathrm{n}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \mathrm{k}−\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \mathrm{k}^{\mathrm{2}} \\ $$$$=\mathrm{n}\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}}−\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)}{\mathrm{6}}\:=\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}}\left\{\mathrm{n}−\frac{\mathrm{2n}+\mathrm{1}}{\mathrm{3}}\right\} \\ $$$$=\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}}\left(\frac{\mathrm{3n}−\mathrm{2n}−\mathrm{1}}{\mathrm{3}}\right)\:=\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}−\mathrm{1}\right)}{\mathrm{6}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com