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AlgebraQuestion and Answers: Page 239

Question Number 119107 Answers: 1 Comments: 0

Question Number 119059 Answers: 1 Comments: 0

Question Number 119055 Answers: 2 Comments: 0

Show by recurence that: ∀ n∈N^∗ Σ_(k=1) ^n (1/(k(k+1)(k+2)))=((n(n+3))/(4(n+1)(n+2)))

$$ \\ $$$${Show}\:{by}\:{recurence}\:{that}:\:\forall\:{n}\in\mathbb{N}^{\ast} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)}=\frac{{n}\left({n}+\mathrm{3}\right)}{\mathrm{4}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)} \\ $$

Question Number 119052 Answers: 3 Comments: 0

Question Number 118967 Answers: 0 Comments: 0

lim_(x→∞) ((((x!)/x^x ))^(1/x) )^(((2∙4)/(1∙3))∙((4∙6)/(3∙5))∙∙∙) =? (2/1)∙(2/3)∙(4/3)∙(4/5)∙(6/5)∙(6/7)∙∙∙=(π/2) lim_(x→∞) ((((x!)/x^x ))^(1/x) )^((2/1)∙(4/3)∙(4/3)∙(6/5)∙∙∙) =((1/e))^(π/2) =(√(((1/e))^π ))=(1/( (√e^π )))

Question Number 118937 Answers: 1 Comments: 0

What is the general rule for factorising (a+b+c)^n ?

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{general}\:\mathrm{rule}\: \\ $$$$\mathrm{for}\:\mathrm{factorising}\:\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)^{\mathrm{n}} ? \\ $$

Question Number 118930 Answers: 1 Comments: 0

Let [x] denote the greatest integer ≤x. Then the number of ordered pair (x,y), where x and y are positive integers less than 30 such that [(x/2)]+[((2x)/3)]+[(y/4)]+[((4y)/5)]=((7x)/6)+((21y)/(20)) is (A) 1 (B) 2 (C) 3 (D) 4

$$\mathrm{Let}\:\left[{x}\right]\:\mathrm{denote}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{integer}\:\leqslant{x}.\:\mathrm{Then}\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{ordered}\:\mathrm{pair}\:\left({x},\mathrm{y}\right),\:\mathrm{where}\:{x}\:\mathrm{and}\:\mathrm{y}\:\mathrm{are} \\ $$$$\mathrm{positive}\:\mathrm{integers}\:\mathrm{less}\:\mathrm{than}\:\mathrm{30}\:\mathrm{such}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\left[\frac{{x}}{\mathrm{2}}\right]+\left[\frac{\mathrm{2}{x}}{\mathrm{3}}\right]+\left[\frac{\mathrm{y}}{\mathrm{4}}\right]+\left[\frac{\mathrm{4y}}{\mathrm{5}}\right]=\frac{\mathrm{7}{x}}{\mathrm{6}}+\frac{\mathrm{21y}}{\mathrm{20}} \\ $$$$\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{4} \\ $$

Question Number 118936 Answers: 3 Comments: 0

show by recurrence that: for n ∈ N^∗ , 2^(6n−5) +3^(2n ) is divisible by 11.

$${show}\:{by}\:{recurrence}\:{that}: \\ $$$${for}\:{n}\:\in\:\mathbb{N}^{\ast} ,\:\mathrm{2}^{\mathrm{6}{n}−\mathrm{5}} +\mathrm{3}^{\mathrm{2}{n}\:} \:{is}\:{divisible}\:{by} \\ $$$$\mathrm{11}. \\ $$

Question Number 118907 Answers: 0 Comments: 0

where can u fond a formula for denesting ramanujan type roots?

$${where}\:{can}\:{u}\:{fond}\:{a}\:{formula}\:{for} \\ $$$${denesting}\:{ramanujan}\:{type}\:{roots}? \\ $$

Question Number 118896 Answers: 1 Comments: 0

Question Number 118849 Answers: 1 Comments: 2

Question Number 118790 Answers: 0 Comments: 0

Show by recurence that (a+b)^n =Σ_(k=0 ) ^n C_n ^k ×a^k ×b^(n−k)

$$\mathrm{Show}\:\mathrm{by}\:\mathrm{recurence}\:\mathrm{that} \\ $$$$\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{n}} =\underset{\mathrm{k}=\mathrm{0}\:} {\overset{\mathrm{n}} {\sum}}\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} ×\mathrm{a}^{\mathrm{k}} ×\mathrm{b}^{\mathrm{n}−\mathrm{k}} \\ $$

Question Number 118780 Answers: 2 Comments: 0

z and z′ ∈ C . show that: 1. zz′^(−) =z^− ×z′^(−) 2. ((z/(z′)))^(−) =(z^− /(z′^(−) ))

$$\mathrm{z}\:\mathrm{and}\:\mathrm{z}'\:\in\:\mathbb{C}\:. \\ $$$$\mathrm{show}\:\mathrm{that}: \\ $$$$\mathrm{1}.\:\:\:\:\:\:\overline {\mathrm{zz}'}=\overset{−} {\mathrm{z}}×\overline {\mathrm{z}'} \\ $$$$\mathrm{2}.\:\:\:\:\:\:\:\overline {\left(\frac{\mathrm{z}}{\mathrm{z}'}\right)}=\frac{\overset{−} {\mathrm{z}}}{\overline {\mathrm{z}'}} \\ $$$$ \\ $$

Question Number 118777 Answers: 1 Comments: 0

lim_(n→∞) (((n!)/n^n ))^(1/n) =lim_(n→∞) ((((√(2nπ)) n^n )/(n^n ×e^n )))^(1/n) ⇒lim_(n→∞) (1/e)((√(2nπ)))^(1/n) =(1/e)

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{n}!}{{n}^{{n}} }\right)^{\frac{\mathrm{1}}{{n}}} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\sqrt{\mathrm{2}{n}\pi}\:{n}^{{n}} }{{n}^{{n}} ×{e}^{{n}} }\right)^{\frac{\mathrm{1}}{{n}}} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{e}}\left(\sqrt{\mathrm{2}{n}\pi}\right)^{\frac{\mathrm{1}}{{n}}} =\frac{\mathrm{1}}{{e}} \\ $$

Question Number 118740 Answers: 1 Comments: 1

f(x+2)+f(x−1)=2x^2 +14 f(x)=?

$${f}\left({x}+\mathrm{2}\right)+{f}\left({x}−\mathrm{1}\right)=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{14} \\ $$$${f}\left({x}\right)=? \\ $$

Question Number 118712 Answers: 4 Comments: 0

Prove the following inequalities: 1)(((n+1)/2))^n >n! for ∀n∈N^∗ ,n>1 2)∣sinnx∣≤n∣sinx∣ for ∀n∈N^∗

$$\mathrm{Prove}\:\mathrm{the}\:\mathrm{following}\:\mathrm{inequalities}: \\ $$$$\left.\mathrm{1}\right)\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{n}} >\mathrm{n}!\:\mathrm{for}\:\forall\mathrm{n}\in\mathrm{N}^{\ast} ,\mathrm{n}>\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\mid\mathrm{sinnx}\mid\leqslant\mathrm{n}\mid\mathrm{sinx}\mid\:\mathrm{for}\:\forall\mathrm{n}\in\mathrm{N}^{\ast} \\ $$

Question Number 118651 Answers: 1 Comments: 2

Question Number 118636 Answers: 1 Comments: 2

Question Number 118634 Answers: 0 Comments: 3

Prove that the equation of the circle passing through the points of intersection of these two curves: y=1+(c/x) ; y=x^2 (c < (2/(3(√3))) ) is (x−(c/2))^2 +(y−1)^2 =1+(c^2 /4) .

$${Prove}\:{that}\:{the}\:{equation}\:{of}\:{the}\:{circle} \\ $$$${passing}\:{through}\:{the}\:{points}\:{of} \\ $$$${intersection}\:{of}\:{these}\:{two}\:{curves}: \\ $$$$\:\:{y}=\mathrm{1}+\frac{{c}}{{x}}\:;\:\:{y}={x}^{\mathrm{2}} \:\:\:\:\:\left({c}\:<\:\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\:\right)\: \\ $$$${is}\:\:\:\left({x}−\frac{{c}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1}+\frac{{c}^{\mathrm{2}} }{\mathrm{4}}\:\:. \\ $$

Question Number 118482 Answers: 1 Comments: 0

If the tangents at the end of a focal chord of parabola meet the tangent at the vertex in C,D.prove that CD substends a right angle at the focus

$$\mathrm{If}\:\mathrm{the}\:\mathrm{tangents}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{end}\:\mathrm{of}\:\mathrm{a}\:\mathrm{focal}\:\:\mathrm{chord}\:\mathrm{of} \\ $$$$\mathrm{parabola}\:\mathrm{meet}\:\mathrm{the} \\ $$$$\mathrm{tangent}\:\mathrm{at}\:\mathrm{the}\:\:\mathrm{vertex} \\ $$$$\mathrm{in}\:\mathrm{C},\mathrm{D}.\mathrm{prove}\:\mathrm{that}\:\mathrm{CD} \\ $$$$\mathrm{substends}\:\mathrm{a}\:\mathrm{right}\:\mathrm{angle} \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{focus} \\ $$

Question Number 118452 Answers: 2 Comments: 3

Question: 2^x +2^(2x+1) +1=y^2 solve this equation if x,y𝛆Z

$$\boldsymbol{{Question}}: \\ $$$$\mathrm{2}^{\boldsymbol{{x}}} +\mathrm{2}^{\mathrm{2}\boldsymbol{{x}}+\mathrm{1}} +\mathrm{1}=\boldsymbol{{y}}^{\mathrm{2}} \:\:\:\boldsymbol{{solve}}\:\boldsymbol{{this}}\:\boldsymbol{{equation}}\:\boldsymbol{{if}} \\ $$$$\boldsymbol{{x}},\boldsymbol{{y}\epsilon}\mathbb{Z} \\ $$

Question Number 118435 Answers: 2 Comments: 0

If 4 (x^9 )^(1/(4 )) −9 (x^9 )^(1/(8 )) + 4 = 0 , then (x^9 )^(1/(4 )) + (x^(−9) )^(1/(4 )) =?

$${If}\:\mathrm{4}\:\sqrt[{\mathrm{4}\:}]{{x}^{\mathrm{9}} }\:−\mathrm{9}\:\sqrt[{\mathrm{8}\:}]{{x}^{\mathrm{9}} }\:+\:\mathrm{4}\:=\:\mathrm{0}\:,\:{then}\: \\ $$$$\:\sqrt[{\mathrm{4}\:}]{{x}^{\mathrm{9}} }\:+\:\sqrt[{\mathrm{4}\:}]{{x}^{−\mathrm{9}} }\:=? \\ $$

Question Number 118391 Answers: 2 Comments: 0

show that if ^ a^2 +b^2 can be divised by 7, a+b can also be divised by 7.

$${show}\:{that}\:{if}\:\:^{} {a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:{can}\:{be}\:{divised} \\ $$$${by}\:\mathrm{7},\:{a}+{b}\:{can}\:{also}\:{be}\:{divised}\:{by}\:\mathrm{7}. \\ $$

Question Number 118371 Answers: 0 Comments: 1

old problem question 118120 tan tan x =tan 3x −tan 2x let t=tan x (1) tan t =((t^5 +2t^3 +t)/(3t^4 −4t^2 +1)) for t≥0 we get (approximating) t_0 =0 t_1 ≈1.28941477 t_2 ≈4.17629616 t_3 ≈7.49316173 t_4 ≈10.7303610 t_5 ≈13.9285293 ... x=nπ+arctan t let n=0 to stay in the first period 0≤t<+∞ ⇒ 0≤arctan t <(π/2) ⇒ (1) has infinite solutions for 0≤x<(π/2) graphically this is easy to see, plot these: f_1 (t)=tan t f_2 (t)=((t(t^4 +2t^2 +1))/(3t^4 −4t^2 +1))=(t/3)+((2t(5t^2 +1))/(3(3t^4 −4t^2 +1))) ⇒ g(t)=(1/3)t is asymptote of f_1 (t) and obviously tan t =at with a∈R has infinite solutions

$$\mathrm{old}\:\mathrm{problem}\:{question}\:\mathrm{118120} \\ $$$$\mathrm{tan}\:\mathrm{tan}\:{x}\:=\mathrm{tan}\:\mathrm{3}{x}\:−\mathrm{tan}\:\mathrm{2}{x} \\ $$$$\mathrm{let}\:{t}=\mathrm{tan}\:{x} \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\mathrm{tan}\:{t}\:=\frac{{t}^{\mathrm{5}} +\mathrm{2}{t}^{\mathrm{3}} +{t}}{\mathrm{3}{t}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{for}\:{t}\geqslant\mathrm{0}\:\mathrm{we}\:\mathrm{get}\:\left(\mathrm{approximating}\right) \\ $$$${t}_{\mathrm{0}} =\mathrm{0} \\ $$$${t}_{\mathrm{1}} \approx\mathrm{1}.\mathrm{28941477} \\ $$$${t}_{\mathrm{2}} \approx\mathrm{4}.\mathrm{17629616} \\ $$$${t}_{\mathrm{3}} \approx\mathrm{7}.\mathrm{49316173} \\ $$$${t}_{\mathrm{4}} \approx\mathrm{10}.\mathrm{7303610} \\ $$$${t}_{\mathrm{5}} \approx\mathrm{13}.\mathrm{9285293} \\ $$$$... \\ $$$${x}={n}\pi+\mathrm{arctan}\:{t} \\ $$$$\mathrm{let}\:{n}=\mathrm{0}\:\mathrm{to}\:\mathrm{stay}\:\mathrm{in}\:\mathrm{the}\:\mathrm{first}\:\mathrm{period} \\ $$$$\mathrm{0}\leqslant{t}<+\infty\:\Rightarrow\:\mathrm{0}\leqslant\mathrm{arctan}\:{t}\:<\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\:\left(\mathrm{1}\right)\:\mathrm{has}\:\mathrm{infinite}\:\mathrm{solutions}\:\mathrm{for}\:\mathrm{0}\leqslant{x}<\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{graphically}\:\mathrm{this}\:\mathrm{is}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{see},\:\mathrm{plot}\:\mathrm{these}: \\ $$$${f}_{\mathrm{1}} \left({t}\right)=\mathrm{tan}\:{t} \\ $$$${f}_{\mathrm{2}} \left({t}\right)=\frac{{t}\left({t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{3}{t}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}=\frac{{t}}{\mathrm{3}}+\frac{\mathrm{2}{t}\left(\mathrm{5}{t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{3}\left(\mathrm{3}{t}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\Rightarrow\:{g}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{3}}{t}\:\mathrm{is}\:\mathrm{asymptote}\:\mathrm{of}\:{f}_{\mathrm{1}} \left({t}\right) \\ $$$$\mathrm{and}\:\mathrm{obviously}\:\mathrm{tan}\:{t}\:={at}\:\mathrm{with}\:{a}\in\mathbb{R}\:\mathrm{has} \\ $$$$\mathrm{infinite}\:\mathrm{solutions} \\ $$

Question Number 118346 Answers: 0 Comments: 0

Let 30 furniture sets arrive at two city stations A and B ,15 sets for each station All funiture sets need to be delivered to two furniture stores C and D,and 10 sets must be delivered to store C,and to store D−20.It is known that delivery one furniture set from the station A to the stores C and D costs 1 and 3 monetary units,and from station B−2 and 5 units respectively,.It is necessary to draw out such a transportation plan that the cost of transportation is the lowest

$$\mathrm{Let}\:\mathrm{30}\:\mathrm{furniture}\:\mathrm{sets}\:\mathrm{arrive}\:\mathrm{at}\:\mathrm{two}\:\mathrm{city} \\ $$$$\mathrm{stations}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:,\mathrm{15}\:\mathrm{sets}\:\mathrm{for}\:\mathrm{each}\:\mathrm{station} \\ $$$$\mathrm{All}\:\mathrm{funiture}\:\mathrm{sets}\:\mathrm{need}\:\mathrm{to}\:\mathrm{be}\:\mathrm{delivered} \\ $$$$\mathrm{to}\:\mathrm{two}\:\mathrm{furniture}\:\mathrm{stores}\:\mathrm{C}\:\mathrm{and}\:\mathrm{D},\mathrm{and}\:\mathrm{10} \\ $$$$\mathrm{sets}\:\mathrm{must}\:\mathrm{be}\:\mathrm{delivered}\:\mathrm{to}\:\mathrm{store}\:\mathrm{C},\mathrm{and} \\ $$$$\mathrm{to}\:\mathrm{store}\:\mathrm{D}−\mathrm{20}.\mathrm{It}\:\mathrm{is}\:\mathrm{known}\:\mathrm{that}\:\mathrm{delivery} \\ $$$$\mathrm{one}\:\mathrm{furniture}\:\mathrm{set}\:\mathrm{from}\:\mathrm{the}\:\mathrm{station}\:\mathrm{A}\:\mathrm{to} \\ $$$$\mathrm{the}\:\mathrm{stores}\:\mathrm{C}\:\mathrm{and}\:\mathrm{D}\:\mathrm{costs}\:\mathrm{1}\:\mathrm{and}\:\mathrm{3}\:\mathrm{monetary} \\ $$$$\mathrm{units},\mathrm{and}\:\mathrm{from}\:\mathrm{station}\:\mathrm{B}−\mathrm{2}\:\mathrm{and}\:\mathrm{5}\:\mathrm{units} \\ $$$$\:\mathrm{respectively},.\mathrm{It}\:\mathrm{is}\:\mathrm{necessary}\:\mathrm{to}\:\mathrm{draw} \\ $$$$\mathrm{out}\:\mathrm{such}\:\mathrm{a}\:\mathrm{transportation}\:\mathrm{plan}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{cost}\:\mathrm{of}\:\mathrm{transportation}\:\mathrm{is}\:\mathrm{the}\:\mathrm{lowest} \\ $$

Question Number 118322 Answers: 2 Comments: 2

solve (1/x)+(1/y)+(1/z)=(3/4) with x,y,z∈N

$${solve} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${with}\:{x},{y},{z}\in\mathbb{N} \\ $$

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