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AlgebraQuestion and Answers: Page 331

Question Number 31039 Answers: 0 Comments: 0

find the sum s_0 = C_n ^o +C_n ^4 +C_n ^8 +... s_1 = C_n ^1 +C_n ^5 +C_n ^9 +.... s_3 = C_n ^2 +C_n ^6 + C_n ^(10) +....

$f i n d t h e s u m$ $s_{0} = C_{n}^{o} + C_{n}^{4} + C_{n}^{8} + . . .$ $s_{1} = C_{n}^{1} + C_{n}^{5} + C_{n}^{9} + . . . .$ $s_{3} = C_{n}^{2} + C_{n}^{6} + C_{n}^{10} + . . . .$

Question Number 31038 Answers: 0 Comments: 0

let give α ∈]−π ,π[ 1)prove that sin^2 α −2(1+cosα) =−4cos^4 ((α/2)) 2)solve inside C z^2 −2z sinα +2(1+cosα)=0 find the module and arg of the roots.

$l e t g i v e α \in] - π, π [$ $1) p r o v e t h a t {s i n}^{2} α - 2 (1 + c o s α) = - 4 {c o s}^{4} (\frac{α}{2})$ $2) s o l v e i n s i d e C z^{2} - 2 z s i n α + 2 (1 + c o s α) = 0 f i n d$ $t h e m o d u l e a n d a r g o f t h e r o o t s .$

Question Number 31037 Answers: 0 Comments: 0

let θ∈[0,π] and Z=1+cosθ +isinθ Z^′ =cosθ +(1+sinθ)i find ∣ZZ^′ ∣ arg(ZZ^′ ) ∣(Z/Z^′ )∣ , arg((Z/Z^, )) , ∣Z−Z^, ∣ and arg(Z−Z^, ).

$l e t θ \in [0, π] a n d Z = 1 + c o s θ + i s i n θ$ $Z^{^{'}} = c o s θ + (1 + s i n θ) i f i n d ∣ {Z Z}^{^{'}} ∣ a r g ({Z Z}^{^{'}})$ $∣ \frac{Z}{Z^{^{'}}} ∣, a r g (\frac{Z}{Z^{,}}), ∣ Z - Z^{,} ∣ a n d a r g (Z - Z^{,}) .$

Question Number 31036 Answers: 0 Comments: 0

solve inside C z^6 = (z^− )^2 .

$s o l v e i n s i d e C z^{6} = {(z^{-})}^{2} .$

Question Number 31035 Answers: 0 Comments: 0

1) solve x^5 =1 2) if x_i are roots of this equation prove that Σ_i x_i =0 3)prove that cos(((2π)/5)) +cos(((4π)/5)) =−(1/2) 4)calculate cos(((4π)/5)) interms cos(((2π)/5)) then find its values.

$1) s o l v e x^{5} = 1$ $2) i f x_{i} a r e r o o t s o f t h i s e q u a t i o n p r o v e t h a t \sum_{i} x_{i} = 0$ $3) p r o v e t h a t c o s (\frac{2 π}{5}) + c o s (\frac{4 π}{5}) = - \frac{1}{2}$ $4) c a l c u l a t e c o s (\frac{4 π}{5}) i n t e r m s c o s (\frac{2 π}{5}) t h e n f i n d i t s$ $v a l u e s .$

Question Number 31034 Answers: 0 Comments: 0

1) solve inside C z^4 =6 2)sove inside C (((z+i)/(z−i)))^3 +(((z+i)/(z−i)))^2 +(((z+i)/(z−i))) +1=0

$1) s o l v e i n s i d e C z^{4} = 6$ $2) s o v e i n s i d e C {(\frac{z + i}{z - i})}^{3} + {(\frac{z + i}{z - i})}^{2} + (\frac{z + i}{z - i}) + 1 = 0$

Question Number 31033 Answers: 0 Comments: 0

factorize p(x)=(1+ix)^n −e^(inθ) wth θ∈R.

$f a c t o r i z e p (x) = {(1 + i x)}^{n} - e^{i n θ} w t h θ \in R .$

Question Number 31032 Answers: 0 Comments: 0

let give x_0 =0 ,y_0 =1 and {_(y_n =x_(n−1) +y_(n−1) ) ^(x_n =x_(n−1) −y_(n−1) ) for n≥1 let z_n =x_n +i y_n ∀n∈N 1)calculate z_0 ,z_1 and z_2 2)prove that ∀n∈N^ ,n≥1 z_n =(1+i)z_(n−1) find z_n then find the expression of x_n and y_n 3)let put S_n =z_0 +z_1 +....z_n s_n =x_0 +x_1 +...+x_n s_n ^′ =y_0 +y_1 +...+y_n find those sum interms of n.

$l e t g i v e x_{0} = 0, y_{0} = 1 a n d {_{y_{n} = x_{n - 1} + y_{n - 1}}^{x_{n} = x_{n - 1} - y_{n - 1}} f o r n ⩾ 1 l e t$ $z_{n} = x_{n} + i y_{n} \forall n \in N$ $1) c a l c u l a t e z_{0}, z_{1} a n d z_{2}$ $2) p r o v e t h a t \forall n \in N^{}, n ⩾ 1 z_{n} = (1 + i) z_{n - 1} f i n d z_{n} t h e n$ $f i n d t h e e x p r e s s i o n o f x_{n} a n d y_{n}$ $3) l e t p u t S_{n} = z_{0} + z_{1} + . . . . z_{n}$ $s_{n} = x_{0} + x_{1} + . . . + x_{n}$ $s_{n}^{^{'}} = y_{0} + y_{1} + . . . + y_{n} f i n d t h o s e s u m i n t e r m s o f n .$

Question Number 31031 Answers: 0 Comments: 0

1) solve inside C z^(12) =1 and give the solution at form r e^(iθ) 2)calculate 1+u +u^2 +... +u^n then find the solution of z∈C z^8 +z^4 +1=0

$1) s o l v e i n s i d e C z^{12} = 1 a n d g i v e t h e s o l u t i o n a t f o r m$ $r e^{i θ}$ $2) c a l c u l a t e 1 + u + u^{2} + . . . + u^{n} t h e n f i n d t h e s o l u t i o n$ $o f z \in C z^{8} + z^{4} + 1 = 0$

Question Number 31023 Answers: 0 Comments: 0

Question Number 31018 Answers: 0 Comments: 1

Question Number 31003 Answers: 1 Comments: 0

Number of positive integers x for which f(x)=x^3 −8x^2 +20x−13 is a prime number are ?

$N u m b e r o f p o s i t i v e i n t e g e r s x f o r$ $w h i c h f (x) = x^{3} - 8 x^{2} + 20 x - 13 i s a$ $p r i m e n u m b e r a r e ?$

Question Number 30990 Answers: 1 Comments: 0

Question Number 30958 Answers: 1 Comments: 4

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.

$S = 3 (1!) - 4 (2!) + 5 (3!) - 6 (4!) + . . . .$ $. . . . . - (2008) (2006!) + 2007!$ $F i n d v a l u e o f S .$

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

$x^{7} + x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1 = 0$ $\underset{k = 1}{\sum^{7}} {[ℜ (x_{k})]}^{2} = ?$ $x_{k} = k^{th} root of the equation$ $ℜ (x_{k}) = real part of the root$

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 .

$l e t U_{n} = {z \in C / z^{n} = 1} s i m l i f y$ $A_{n} = \sum_{α \in U_{n}} {(x + α)}^{n} a n d B_{n} = \sum_{α \in U_{n}} {(x - α)}^{n} .$

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.

$l e t p (x) = {(x - 1)}^{n} - x^{n} + 1 w i t h n i n t e g r f i n d n$ $i n o r d r e t h a t p (x) h a v e a d o u b l e r o o t .$

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))) .

$d e c o m p o s e i n s i d e R [x] p (x) = x^{2 n + 1} - 1 t h e n f i n d$ $\prod_{k = 1}^{n} s i n (\frac{k π}{2 n + 1}) .$

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)) .

$p r o v e t h a t \forall p \in N i t e x i s t o n e p o l y n o m i a l Q_{2 p} /$ $s i n (2 p + 1) θ = {s i n}^{2 p + 1} θ Q_{2 p} (c o t a n θ) a n d {d e g Q}_{2 p} = 2 p$ $2) p r o v e t h a t \prod_{k = 1}^{p} t a n (\frac{k π}{2 p + 1}) = \sqrt{2 p + 1} .$

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)).

$l e t (u_{n}) / u_{1} = 1 - i a n d \forall p \in {2, 3, . . . n} u_{p} = u_{p - 1} j w i t h$ $j = e^{i \frac{2 π}{3}}$ $1) v e r i f y t h a t u_{1} + u_{2} + u_{3} = 0$ $2) p r o v e t h a t \forall p \in {4, 5, . . ., n} u_{p} = u_{p - 3}$ $3) f i n d t h e v a l u e o f S_{n} = \sum_{i = 1}^{n} u_{i}$ $4) c a l c u l a t e α_{n} = \sum_{p = 0}^{n - 1} c o s (- \frac{π}{4} + \frac{2 p π}{3}) a n d$ $β_{n} = \sum_{p = 0}^{n - 1} s i n (- \frac{π}{4} + \frac{2 p π}{3}) .$

Question Number 30627 Answers: 1 Comments: 0

Question Number 30600 Answers: 0 Comments: 0

let w_k =e^(i((2kπ)/n)) find A= Π_(k=0) ^(n−1) (a +bw_k ).

$l e t w_{k} = e^{i \frac{2 k π}{n}} f i n d A = \prod_{k = 0}^{n - 1} (a + {b w}_{k}) .$

Question Number 30599 Answers: 0 Comments: 1

decompose inside C(x) F= (1/((x−1)(x^n −1))) .

$d e c o m p o s e i n s i d e C (x) F = \frac{1}{(x - 1) (x^{n} - 1)} .$

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