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AlgebraQuestion and Answers: Page 327

Question Number 38459 Answers: 0 Comments: 7

x^2 (x−b^2 )+a^2 b^2 (x−a^2 )=0 Solve for x.

$${x}^{\mathrm{2}} \left({x}−{b}^{\mathrm{2}} \right)+{a}^{\mathrm{2}} {b}^{\mathrm{2}} \left({x}−{a}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${Solve}\:{for}\:{x}. \\ $$

Question Number 37991 Answers: 1 Comments: 4

1. Find the sum s_n =1+2x+3x^2 +4x^3 +...+nx^(n−1) Hence,or otherwise, find the sum Σ_(k=1) ^n k.2^k 2. Simplify the following i. Σ_(r=0) ^n (_(2r−1) ^(2n) ) ii.Σ_(r=0) ^n (−1)^r r(_r ^n ) iii.Σ_(r=0) ^n (−1)^r (1/(r+1))(_r ^n ) iv.Σ_(r=0) ^n (_(2r) ^(2n) ) v.Σ_(r=0) ^n (−1)^r (_(n−r) ^(n+1) ) 3.Find the sum Σ_(r=0) ^(n−k) (_k ^(n−r) ), where k=0,1,2,3,...,n

$$\mathrm{1}.\:{Find}\:{the}\:{sum} \\ $$$$\:\:\:\:{s}_{{n}} =\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{3}} +...+{nx}^{{n}−\mathrm{1}} \\ $$$${Hence},{or}\:{otherwise},\:{find}\:{the}\:{sum} \\ $$$$\:\:\:\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}.\mathrm{2}^{{k}} \\ $$$$\mathrm{2}.\:{Simplify}\:{the}\:{following} \\ $$$${i}.\:\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\left(_{\mathrm{2}{r}−\mathrm{1}} ^{\mathrm{2}{n}} \right) \\ $$$${ii}.\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{r}} {r}\left(_{{r}} ^{{n}} \right) \\ $$$${iii}.\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{r}} \frac{\mathrm{1}}{{r}+\mathrm{1}}\left(_{{r}} ^{{n}} \right) \\ $$$${iv}.\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\left(_{\mathrm{2}{r}} ^{\mathrm{2}{n}} \right) \\ $$$${v}.\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{r}} \left(_{{n}−{r}} ^{{n}+\mathrm{1}} \right) \\ $$$$\mathrm{3}.{Find}\:{the}\:{sum} \\ $$$$\:\:\:\:\:\:\:\:\:\underset{{r}=\mathrm{0}} {\overset{{n}−{k}} {\sum}}\left(_{{k}} ^{{n}−{r}} \right),\:\:\:{where}\:{k}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},...,{n} \\ $$

Question Number 37953 Answers: 0 Comments: 1

Question Number 37906 Answers: 1 Comments: 0

Question Number 37890 Answers: 0 Comments: 2

calculate A_n = Σ_(k=0) ^n (−1)^k (2k+3) interms of n

$${calculate}\:{A}_{{n}} =\:\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \left(\mathrm{2}{k}+\mathrm{3}\right)\:{interms}\:{of}\:{n} \\ $$

Question Number 37885 Answers: 0 Comments: 0

find Π_(k=1) ^n cos(((kπ)/(2n+1))) and Π_(k=1) ^n sin(((kπ)/(2n+1)))

$${find}\:\:\:\prod_{{k}=\mathrm{1}} ^{{n}} \:{cos}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\:\:{and}\:\prod_{{k}=\mathrm{1}} ^{{n}} \:{sin}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right) \\ $$

Question Number 37642 Answers: 1 Comments: 0

3^(3(√(250))+7^(3(√(16))−4^(3(√(54))) ) )

$$\mathrm{3}^{\mathrm{3}\sqrt{\mathrm{250}}+\mathrm{7}^{\mathrm{3}\sqrt{\mathrm{16}}−\mathrm{4}^{\mathrm{3}\sqrt{\mathrm{54}}} } } \\ $$

Question Number 37627 Answers: 3 Comments: 0

Given {x} = x − ⌊x⌋ How many real solutions from equation {x} + {x^2 } = 1 with −10 ≤ x ≤ 10 ?

$$\mathrm{Given}\:\left\{{x}\right\}\:=\:{x}\:−\:\lfloor{x}\rfloor \\ $$$$\mathrm{How}\:\mathrm{many}\:\mathrm{real}\:\mathrm{solutions}\:\mathrm{from}\:\mathrm{equation} \\ $$$$\left\{{x}\right\}\:+\:\left\{{x}^{\mathrm{2}} \right\}\:=\:\mathrm{1} \\ $$$$\mathrm{with}\:−\mathrm{10}\:\leqslant\:{x}\:\leqslant\:\mathrm{10}\:? \\ $$

Question Number 37568 Answers: 2 Comments: 0

let α and β the roots of the equation x^2 −2mx −1 =0 find interms of the real m A = α^2 +β^2 B =α^3 +β^3 c =α^4 +β^4 D= α^6 +β^6

$${let}\:\alpha\:{and}\:\beta\:{the}\:{roots}\:{of}\:{the}\:{equation} \\ $$$${x}^{\mathrm{2}} \:−\mathrm{2}{mx}\:−\mathrm{1}\:=\mathrm{0}\:\:{find}\:{interms}\:{of}\:{the}\:{real}\:{m} \\ $$$${A}\:=\:\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} \\ $$$${B}\:=\alpha^{\mathrm{3}} \:+\beta^{\mathrm{3}} \\ $$$${c}\:=\alpha^{\mathrm{4}} \:+\beta^{\mathrm{4}} \\ $$$${D}=\:\alpha^{\mathrm{6}} \:+\beta^{\mathrm{6}} \\ $$

Question Number 37464 Answers: 1 Comments: 2

36x^2 y^2 -42x^3 y^3 +24x^5 y^4

$$\mathrm{36}\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{y}}^{\mathrm{2}} -\mathrm{42}\boldsymbol{{x}}^{\mathrm{3}} \boldsymbol{{y}}^{\mathrm{3}} +\mathrm{24}\boldsymbol{{x}}^{\mathrm{5}} \boldsymbol{{y}}^{\mathrm{4}} \\ $$

Question Number 37414 Answers: 1 Comments: 0

x^x =2 find the value of x.

$${x}^{{x}} =\mathrm{2} \\ $$$$ \\ $$$${find}\:{the}\:{value}\:{of}\:{x}. \\ $$

Question Number 37411 Answers: 1 Comments: 0

((√(2)^(√2) ))

$$\left(\sqrt{\left.\mathrm{2}\right)^{\sqrt{\mathrm{2}}} }\right. \\ $$

Question Number 37324 Answers: 0 Comments: 3

Question Number 37232 Answers: 0 Comments: 0

let A = (((1 −1 0)),((−1 1 1)) ) (0 0 3 ) calculate A^n .

$${let}\:{A}\:=\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\:−\mathrm{1}\:\:\:\:\:\:\:\mathrm{0}}\\{−\mathrm{1}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\mathrm{3}\:\right) \\ $$$${calculate}\:{A}^{{n}} \:. \\ $$

Question Number 37231 Answers: 0 Comments: 0

let A = (((0 1 1)),((1 0 1)) ) (1 1 0 ) 1) calculate p_c (A) the caracteristic polunom of A 2) calculate A^n with n integr natural 3) calcypulate e^(tA) t∈ R

$${let}\:{A}\:=\:\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{0}\:\right) \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{p}_{{c}} \left({A}\right)\:{the}\:{caracteristic}\: \\ $$$${polunom}\:{of}\:{A} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{A}^{{n}} \:\:{with}\:{n}\:{integr}\:{natural} \\ $$$$\left.\mathrm{3}\right)\:{calcypulate}\:{e}^{{tA}} \:\:\:\:\:{t}\in\:{R}\:\: \\ $$

Question Number 37230 Answers: 0 Comments: 0

let A = (((1 −2)),((1 4)) ) calculate A^n 2) find e^A , e^(−A) 3) find e^(iA) , e^(−iA) and e^(iA) +e^(−iA) .

$${let}\:{A}\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:−\mathrm{2}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}}\end{pmatrix} \\ $$$${calculate}\:\:{A}^{{n}} \: \\ $$$$\left.\mathrm{2}\right)\:{find}\:\:{e}^{{A}} \:\:,\:{e}^{−{A}} \\ $$$$\left.\mathrm{3}\right)\:{find}\:\:{e}^{{iA}} ,\:{e}^{−{iA}} \:\:\:{and}\:{e}^{{iA}} \:+{e}^{−{iA}} \:\:. \\ $$

Question Number 37228 Answers: 0 Comments: 0

E id k vectorial space and f∈L(E) 1)prove that if f is nilpotent with indice p≥1 ,I −f is bijective and (I−f)^(−1) =Σ_(i=0) ^(p−1) f^i 2)let E=R_n [x] and f∈L(E) / f(p) =p−p^′ prove that f is inversible and find f^(−1) .

$${E}\:{id}\:{k}\:{vectorial}\:{space}\:{and}\:{f}\in{L}\left({E}\right) \\ $$$$\left.\mathrm{1}\right){prove}\:{that}\:{if}\:{f}\:{is}\:{nilpotent}\:{with}\:{indice} \\ $$$${p}\geqslant\mathrm{1}\:,{I}\:−{f}\:{is}\:{bijective}\:{and} \\ $$$$\left({I}−{f}\right)^{−\mathrm{1}} =\sum_{{i}=\mathrm{0}} ^{{p}−\mathrm{1}} {f}^{{i}} \\ $$$$\left.\mathrm{2}\right){let}\:{E}={R}_{{n}} \left[{x}\right]\:{and}\:{f}\in{L}\left({E}\right)\:/ \\ $$$${f}\left({p}\right)\:={p}−{p}^{'} \:\:{prove}\:{that}\:{f}\:{is}\:{inversible} \\ $$$${and}\:{find}\:{f}^{−\mathrm{1}} \:. \\ $$

Question Number 37166 Answers: 0 Comments: 2

if 3x^2 +2αxy+2y^2 +2ax−4y+1 can be resolved into two linear factors, prove that ′α′ is a root of the equation x^2 +4ax+2a^2 +6=0

$${if}\:\:\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}\alpha{xy}+\mathrm{2}{y}^{\mathrm{2}} +\mathrm{2}{ax}−\mathrm{4}{y}+\mathrm{1} \\ $$$${can}\:{be}\:{resolved}\:\:{into}\:\:{two}\:\:{linear} \\ $$$${factors},\:\:{prove}\:\:{that}\:\:'\alpha'\:\:{is}\:{a}\:{root}\: \\ $$$${of}\:{the}\:{equation}\:{x}^{\mathrm{2}} +\mathrm{4}{ax}+\mathrm{2}{a}^{\mathrm{2}} +\mathrm{6}=\mathrm{0} \\ $$

Question Number 37139 Answers: 2 Comments: 0

if α , β are the roots of the quadratic equation ax^2 +bx+c =0 then find the quadratic equation whose roots are α^(2 ) , β^2

$${if}\:\alpha\:,\:\beta\:\:{are}\:{the}\:{roots}\:{of}\:{the}\:{quadratic} \\ $$$${equation}\:{ax}^{\mathrm{2}} +{bx}+{c}\:=\mathrm{0}\:{then}\:\:{find} \\ $$$${the}\:{quadratic}\:{equation}\:{whose}\:{roots} \\ $$$${are}\:\:\alpha^{\mathrm{2}\:\:\:} ,\:\beta^{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$

Question Number 36926 Answers: 1 Comments: 0

let f(x) = e^(−x^2 ) 1) prove that f^((n)) (x)=p_n (x)e^(−x^2 ) with p_n is a polynom 2) find a relation of recurrence between the p_n 3) calculate p_1 ,p_2 ,p_3 ,p_4

$${let}\:{f}\left({x}\right)\:=\:{e}^{−{x}^{\mathrm{2}} } \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:{f}^{\left({n}\right)} \left({x}\right)={p}_{{n}} \left({x}\right){e}^{−{x}^{\mathrm{2}} } \:\:{with}\:{p}_{{n}} \:{is}\:{a}\:{polynom} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{a}\:{relation}\:{of}\:{recurrence}\:{between}\:{the}\:{p}_{{n}} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:{p}_{\mathrm{1}} ,{p}_{\mathrm{2}} ,{p}_{\mathrm{3}} ,{p}_{\mathrm{4}} \\ $$

Question Number 36911 Answers: 0 Comments: 1

p is a polynome having nroots simples x_i (1≤x_i ≤n ) with x_i ^2 ≠1 calculste Σ_(k=1) ^n (1/(1−x_k )) .

$${p}\:{is}\:{a}\:{polynome}\:{having}\:{nroots}\:{simples} \\ $$$${x}_{{i}} \:\left(\mathrm{1}\leqslant{x}_{{i}} \leqslant{n}\:\right)\:{with}\:{x}_{{i}} ^{\mathrm{2}} \:\neq\mathrm{1}\:\:{calculste} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{1}−{x}_{{k}} }\:. \\ $$

Question Number 36909 Answers: 0 Comments: 0

let p(x)=x^3 −2x^2 −1 and α is root of p(x) prove that α∉ Q .

$${let}\:{p}\left({x}\right)={x}^{\mathrm{3}} \:−\mathrm{2}{x}^{\mathrm{2}} \:−\mathrm{1}\:{and}\:\alpha\:{is}\:{root}\:{of}\:{p}\left({x}\right) \\ $$$${prove}\:{that}\:\alpha\notin\:{Q}\:. \\ $$

Question Number 36905 Answers: 0 Comments: 0

p is apolynom with n roots differents let Q = p^2 +p^′ let α the number of roots of Q prove that n−1≤α≤n+1 .

$${p}\:{is}\:{apolynom}\:{with}\:{n}\:{roots}\:{differents} \\ $$$${let}\:{Q}\:=\:{p}^{\mathrm{2}} \:+{p}^{'} \:\:\:\:{let}\:\alpha\:{the}\:{number}\:{of}\:{roots}\:{of} \\ $$$${Q}\:{prove}\:{that}\:\:\:{n}−\mathrm{1}\leqslant\alpha\leqslant{n}+\mathrm{1}\:. \\ $$

Question Number 36904 Answers: 0 Comments: 1

1)decompose inside C[x] p(x)=x^(2n) −2(cosα)x^n +1 2) decopose p(x)inside R[x]