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Question Number 73039 by mathmax by abdo last updated on 05/Nov/19
let U_n =(n/2) if n even and U_n =((n−1)/2) if n odd let f(n)=Σ_(k=0) ^n U_k prove that ∀(x,y)∈N^2 f(x+y)−f(x−y)=xy
$${let}\:{U}_{{n}} =\frac{{n}}{\mathrm{2}}\:{if}\:{n}\:{even}\:{and}\:{U}_{{n}} =\frac{{n}−\mathrm{1}}{\mathrm{2}}\:{if}\:{n}\:{odd}\:{let}\:{f}\left({n}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} {U}_{{k}} \\ $$$${prove}\:{that}\:\forall\left({x},{y}\right)\in{N}^{\mathrm{2}} \:\:\:\:{f}\left({x}+{y}\right)−{f}\left({x}−{y}\right)={xy} \\ $$
Answered by mind is power last updated on 05/Nov/19
just for the definition of f x≥y x+y,x−y has sam parite f(n)=Σ_(k=0) ^n U_k =Σ_(k=0) ^(E((n/2))) k+Σ_(k=1) ^(E(((n−1)/2))) k=(((E((n/2))+1).(E((n/2))))/2)+((E(((n−1)/2)).(1+E(((n−1)/2))))/2) f(x+y)−f(x−y) is x+y=2n x−y=2n−2y ⇒f(x+y)−f(x−y)=((n(n+1))/2)−(((n−y+1)(n−y))/2)+(((n−1)(n))/2)−(((n−y−1)(n−y))/2) =((n^2 +n−(n^2 −2ny+y^2 +n−y)+n^2 −n−(n^2 +y^2 −2ny−n+y))/2) =((−2y^2 +4ny)/2)=((−2y^2 +2(x+y)y)/2)=xy same thing if x+y=2n+1
$$\mathrm{just}\:\mathrm{for}\:\mathrm{the}\:\mathrm{definition}\:\mathrm{of}\:\mathrm{f}\: \\ $$$$\mathrm{x}\geqslant\mathrm{y} \\ $$$$\mathrm{x}+\mathrm{y},\mathrm{x}−\mathrm{y}\:\mathrm{has}\:\mathrm{sam}\:\mathrm{parite} \\ $$$$\mathrm{f}\left(\mathrm{n}\right)=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \mathrm{U}_{\mathrm{k}} =\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{E}\left(\frac{\mathrm{n}}{\mathrm{2}}\right)} {\sum}}\mathrm{k}+\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{E}\left(\frac{\mathrm{n}−\mathrm{1}}{\mathrm{2}}\right)} {\sum}}\mathrm{k}=\frac{\left(\mathrm{E}\left(\frac{\mathrm{n}}{\mathrm{2}}\right)+\mathrm{1}\right).\left(\mathrm{E}\left(\frac{\mathrm{n}}{\mathrm{2}}\right)\right)}{\mathrm{2}}+\frac{\mathrm{E}\left(\frac{\mathrm{n}−\mathrm{1}}{\mathrm{2}}\right).\left(\mathrm{1}+\mathrm{E}\left(\frac{\mathrm{n}−\mathrm{1}}{\mathrm{2}}\right)\right)}{\mathrm{2}} \\ $$$$\mathrm{f}\left(\mathrm{x}+\mathrm{y}\right)−\mathrm{f}\left(\mathrm{x}−\mathrm{y}\right) \\ $$$$\mathrm{is}\:\mathrm{x}+\mathrm{y}=\mathrm{2n} \\ $$$$\mathrm{x}−\mathrm{y}=\mathrm{2n}−\mathrm{2y} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}+\mathrm{y}\right)−\mathrm{f}\left(\mathrm{x}−\mathrm{y}\right)=\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}}−\frac{\left(\mathrm{n}−\mathrm{y}+\mathrm{1}\right)\left(\mathrm{n}−\mathrm{y}\right)}{\mathrm{2}}+\frac{\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}\right)}{\mathrm{2}}−\frac{\left(\mathrm{n}−\mathrm{y}−\mathrm{1}\right)\left(\mathrm{n}−\mathrm{y}\right)}{\mathrm{2}} \\ $$$$=\frac{\mathrm{n}^{\mathrm{2}} +\mathrm{n}−\left(\mathrm{n}^{\mathrm{2}} −\mathrm{2ny}+\mathrm{y}^{\mathrm{2}} +\mathrm{n}−\mathrm{y}\right)+\mathrm{n}^{\mathrm{2}} −\mathrm{n}−\left(\mathrm{n}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{2ny}−\mathrm{n}+\mathrm{y}\right)}{\mathrm{2}} \\ $$$$=\frac{−\mathrm{2y}^{\mathrm{2}} +\mathrm{4ny}}{\mathrm{2}}=\frac{−\mathrm{2y}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{x}+\mathrm{y}\right)\mathrm{y}}{\mathrm{2}}=\mathrm{xy} \\ $$$$\mathrm{same}\:\mathrm{thing}\:\mathrm{if}\:\mathrm{x}+\mathrm{y}=\mathrm{2n}+\mathrm{1} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 05/Nov/19
thanx sir.
$${thanx}\:{sir}. \\ $$

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