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AlgebraQuestion and Answers: Page 327

Question Number 30956 Answers: 1 Comments: 0

Question Number 30916 Answers: 1 Comments: 1

Question Number 30860 Answers: 1 Comments: 0

S= 3(1!)−4(2!)+5(3!)−6(4!)+.... .....−(2008)(2006!)+2007! Find value of S.

$${S}=\:\mathrm{3}\left(\mathrm{1}!\right)−\mathrm{4}\left(\mathrm{2}!\right)+\mathrm{5}\left(\mathrm{3}!\right)−\mathrm{6}\left(\mathrm{4}!\right)+.... \\ $$$$\:\:\:\:.....−\left(\mathrm{2008}\right)\left(\mathrm{2006}!\right)+\mathrm{2007}! \\ $$$${Find}\:{value}\:{of}\:{S}. \\ $$

Question Number 30849 Answers: 0 Comments: 5

x^7 +x^6 +x^5 +x^4 +x^3 +x^2 +x+1=0 Σ_(k=1) ^7 [ℜ(x_k )]^2 = ? x_k = k^( th) root of the equation ℜ(x_k ) = real part of the root

$${x}^{\mathrm{7}} +{x}^{\mathrm{6}} +{x}^{\mathrm{5}} +{x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0} \\ $$$$\: \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{7}} {\sum}}\left[\Re\left({x}_{{k}} \right)\right]^{\mathrm{2}} \:=\:? \\ $$$${x}_{{k}} \:=\:{k}^{\:\mathrm{th}} \:\mathrm{root}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\Re\left({x}_{{k}} \right)\:=\:\mathrm{real}\:\mathrm{part}\:\mathrm{of}\:\mathrm{the}\:\mathrm{root} \\ $$

Question Number 30745 Answers: 0 Comments: 0

let U_n ={z∈C/z^n =1} simlify A_n = Σ_(α∈U_n ) (x+α)^n and B_n =Σ_(α∈ U_n ) (x−α)^n .

$${let}\:\:{U}_{{n}} =\left\{{z}\in{C}/{z}^{{n}} =\mathrm{1}\right\}\:\:{simlify} \\ $$$${A}_{{n}} =\:\sum_{\alpha\in{U}_{{n}} } \:\left({x}+\alpha\right)^{{n}} \:{and}\:{B}_{{n}} =\sum_{\alpha\in\:{U}_{{n}} } \:\:\left({x}−\alpha\right)^{{n}} . \\ $$

Question Number 30744 Answers: 0 Comments: 0

let p(x)= (x−1)^n −x^n +1 with n integr find n in ordre that p(x) have a double root.

$${let}\:{p}\left({x}\right)=\:\left({x}−\mathrm{1}\right)^{{n}} \:−{x}^{{n}} \:+\mathrm{1}\:\:{with}\:{n}\:{integr}\:{find}\:{n} \\ $$$${in}\:{ordre}\:{that}\:{p}\left({x}\right)\:{have}\:{a}\:{double}\:{root}. \\ $$

Question Number 30743 Answers: 0 Comments: 1

decompose inside R[x] p(x)=x^(2n+1) −1 then find Π_(k=1) ^n sin( ((kπ)/(2n+1))) .

$${decompose}\:{inside}\:{R}\left[{x}\right]\:\:{p}\left({x}\right)={x}^{\mathrm{2}{n}+\mathrm{1}} \:−\mathrm{1}\:{then}\:{find} \\ $$$$\prod_{{k}=\mathrm{1}} ^{{n}} \:{sin}\left(\:\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\:. \\ $$

Question Number 30742 Answers: 0 Comments: 0

prove that ∀p ∈N it exist one polynomial Q_(2p) / sin(2p+1)θ=sin^(2p+1) θ Q_(2p) (cotanθ) and degQ_(2p) =2p 2) prove that Π_(k=1) ^p tan(((kπ)/(2p+1)))=(√(2p+1)) .

$${prove}\:{that}\:\forall{p}\:\in{N}\:\:{it}\:{exist}\:{one}\:{polynomial}\:{Q}_{\mathrm{2}{p}} \:/ \\ $$$${sin}\left(\mathrm{2}{p}+\mathrm{1}\right)\theta={sin}^{\mathrm{2}{p}+\mathrm{1}} \theta\:{Q}_{\mathrm{2}{p}} \:\left({cotan}\theta\right)\:{and}\:{degQ}_{\mathrm{2}{p}} =\mathrm{2}{p} \\ $$$$\left.\mathrm{2}\right)\:{prove}\:{that}\:\prod_{{k}=\mathrm{1}} ^{{p}} \:{tan}\left(\frac{{k}\pi}{\mathrm{2}{p}+\mathrm{1}}\right)=\sqrt{\mathrm{2}{p}+\mathrm{1}}\:. \\ $$$$ \\ $$

Question Number 30739 Answers: 0 Comments: 0

let (u_n ) / u_1 =1−i and ∀p∈{2,3,...n} u_p =u_(p−1) j with j=e^(i((2π)/3)) 1)verify that u_1 +u_2 +u_3 =0 2)prove that ∀p∈ {4,5,...,n} u_p =u_(p−3) 3)find the value of S_n =Σ_(i=1) ^n u_i 4)calculate α_n = Σ_(p=0) ^(n−1) cos(−(π/4) +((2pπ)/3)) and β_n = Σ_(p=0) ^(n−1) sin(−(π/4) +((2pπ)/3)).

$${let}\:\left({u}_{{n}} \right)\:/\:{u}_{\mathrm{1}} =\mathrm{1}−{i}\:{and}\:\:\forall{p}\in\left\{\mathrm{2},\mathrm{3},...{n}\right\}\:{u}_{{p}} ={u}_{{p}−\mathrm{1}} {j}\:{with} \\ $$$${j}={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \\ $$$$\left.\mathrm{1}\right){verify}\:{that}\:{u}_{\mathrm{1}} \:+{u}_{\mathrm{2}} \:+{u}_{\mathrm{3}} =\mathrm{0} \\ $$$$\left.\mathrm{2}\right){prove}\:{that}\:\forall{p}\in\:\left\{\mathrm{4},\mathrm{5},...,{n}\right\}\:\:{u}_{{p}} ={u}_{{p}−\mathrm{3}} \\ $$$$\left.\mathrm{3}\right){find}\:{the}\:{value}\:{of}\:{S}_{{n}} \:=\sum_{{i}=\mathrm{1}} ^{{n}} \:{u}_{{i}} \\ $$$$\left.\mathrm{4}\right){calculate}\:\:\alpha_{{n}} =\:\sum_{{p}=\mathrm{0}} ^{{n}−\mathrm{1}} {cos}\left(−\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{2}{p}\pi}{\mathrm{3}}\right)\:{and} \\ $$$$\beta_{{n}} =\:\sum_{{p}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{sin}\left(−\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{2}{p}\pi}{\mathrm{3}}\right). \\ $$

Question Number 30627 Answers: 1 Comments: 0