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Question Number 119635    Answers: 1   Comments: 0

If a function f:R→R satisfies the relation f(x+1)+f(x−1)=(√3)f(x) for all x∈R then a period of f is (A) 10 (B) 12 (C) 6 (D) 4

$$\mathrm{If}\:\mathrm{a}\:\mathrm{function}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{satisfies}\:\mathrm{the}\:\mathrm{relation} \\ $$$$\:\:\:\:\:\:\:{f}\left(\mathrm{x}+\mathrm{1}\right)+{f}\left(\mathrm{x}−\mathrm{1}\right)=\sqrt{\mathrm{3}}{f}\left(\mathrm{x}\right)\:\mathrm{for}\:\mathrm{all}\:\mathrm{x}\in\mathbb{R} \\ $$$$\mathrm{then}\:\mathrm{a}\:\mathrm{period}\:\mathrm{of}\:{f}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{10}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{12}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{6}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{4} \\ $$

Question Number 119634    Answers: 2   Comments: 0

Q1 If f:R→R is defined by f(x)=[x]+[x+(1/2)]+[x+(2/3)]−3x+5 where [x] is the integral part of x, then a period of f is (A) 1 (B) 2/3 (C) 1/2 (D) 1/3 Q2 Let a<c<b such that c−a=b−c. If f:R→R is a function satisfying the relation f(x+a)+f(x+b)=f(x+c) for all x∈R then a period of f is (A) (b−a) (B) 2(b−a) (C) 3(b−a) (D) 4(b−a)

$$\mathrm{Q1} \\ $$$$\mathrm{If}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{by} \\ $$$$\:\:\:\:\:\:\:\:{f}\left(\mathrm{x}\right)=\left[\mathrm{x}\right]+\left[\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right]+\left[\mathrm{x}+\frac{\mathrm{2}}{\mathrm{3}}\right]−\mathrm{3x}+\mathrm{5} \\ $$$$\mathrm{where}\:\left[\mathrm{x}\right]\:\mathrm{is}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{part}\:\mathrm{of}\:\mathrm{x},\:\mathrm{then}\:\mathrm{a}\:\mathrm{period}\:\mathrm{of}\:{f}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{2}/\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{1}/\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{1}/\mathrm{3} \\ $$$$ \\ $$$$\mathrm{Q2} \\ $$$$\mathrm{Let}\:{a}<\mathrm{c}<\mathrm{b}\:\mathrm{such}\:\mathrm{that}\:\mathrm{c}−{a}=\mathrm{b}−\mathrm{c}.\:\mathrm{If}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{function}\:\mathrm{satisfying}\:\mathrm{the}\:\mathrm{relation} \\ $$$$\:\:{f}\left(\mathrm{x}+{a}\right)+{f}\left(\mathrm{x}+\mathrm{b}\right)={f}\left(\mathrm{x}+\mathrm{c}\right)\:\:\mathrm{for}\:\mathrm{all}\:\mathrm{x}\in\mathbb{R} \\ $$$$\mathrm{then}\:\mathrm{a}\:\mathrm{period}\:\mathrm{of}\:{f}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\left(\mathrm{b}−{a}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{2}\left(\mathrm{b}−{a}\right) \\ $$$$\left(\mathrm{C}\right)\:\mathrm{3}\left(\mathrm{b}−{a}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{4}\left(\mathrm{b}−{a}\right) \\ $$

Question Number 119580    Answers: 3   Comments: 0

Given k ∈ N. 1) justify these relations: 3^(2k) +1≡2[8] and 3^(2k+1) +1≡4[8]. 2) Given (E): 2^n −3^m =1. n and m are unknowed. • Show that if m is even , (E) does not have solution. ■ Deduct from the first question 1) that the couple (2;1) is the only solution of (E).

$$\mathrm{Given}\:\mathrm{k}\:\in\:\mathbb{N}. \\ $$$$\left.\mathrm{1}\right)\:\mathrm{justify}\:\mathrm{these}\:\mathrm{relations}: \\ $$$$\mathrm{3}^{\mathrm{2k}} +\mathrm{1}\equiv\mathrm{2}\left[\mathrm{8}\right]\:\mathrm{and}\:\mathrm{3}^{\mathrm{2k}+\mathrm{1}} +\mathrm{1}\equiv\mathrm{4}\left[\mathrm{8}\right]. \\ $$$$\left.\mathrm{2}\right)\:\mathrm{Given}\:\left(\mathrm{E}\right):\:\mathrm{2}^{\mathrm{n}} −\mathrm{3}^{\mathrm{m}} =\mathrm{1}.\:\mathrm{n}\:\mathrm{and}\:\mathrm{m}\:\mathrm{are}\:\mathrm{unknowed}. \\ $$$$\bullet\:\mathrm{Show}\:\mathrm{that}\:\mathrm{if}\:\mathrm{m}\:\mathrm{is}\:\mathrm{even}\:,\:\left(\mathrm{E}\right)\:\mathrm{does}\:\mathrm{not}\:\mathrm{have}\: \\ $$$$\mathrm{solution}. \\ $$$$\left.\blacksquare\:\mathrm{Deduct}\:\mathrm{from}\:\mathrm{the}\:\mathrm{first}\:\mathrm{question}\:\mathrm{1}\right)\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{couple}\:\left(\mathrm{2};\mathrm{1}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{only}\:\mathrm{solution}\:\mathrm{of}\:\left(\mathrm{E}\right). \\ $$

Question Number 119576    Answers: 2   Comments: 0

Question Number 119540    Answers: 1   Comments: 0

Question Number 119538    Answers: 1   Comments: 1

Question Number 119490    Answers: 0   Comments: 0

Let a<c<b such that c−a=b−c. If f:R→R is a function satisfying the relation f(x+a)+f(x+b)=f(x+c) for all x∈R then a period of f is (A) (b−a) (B) 2(b−a) (C) 3(b−a) (D) 4(b−a)

$$\mathrm{Let}\:{a}<\mathrm{c}<\mathrm{b}\:\mathrm{such}\:\mathrm{that}\:\mathrm{c}−{a}=\mathrm{b}−\mathrm{c}.\:\mathrm{If}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{function}\:\mathrm{satisfying}\:\mathrm{the}\:\mathrm{relation} \\ $$$$\:\:{f}\left(\mathrm{x}+{a}\right)+{f}\left(\mathrm{x}+\mathrm{b}\right)={f}\left(\mathrm{x}+\mathrm{c}\right)\:\:\mathrm{for}\:\mathrm{all}\:\mathrm{x}\in\mathbb{R} \\ $$$$\mathrm{then}\:\mathrm{a}\:\mathrm{period}\:\mathrm{of}\:{f}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\left(\mathrm{b}−{a}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{2}\left(\mathrm{b}−{a}\right) \\ $$$$\left(\mathrm{C}\right)\:\mathrm{3}\left(\mathrm{b}−{a}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{4}\left(\mathrm{b}−{a}\right) \\ $$

Question Number 119483    Answers: 1   Comments: 0

Let a>0 and f:R→R a function satisfying f(x+a)=1+[2−3f(x)+3f(x)^2 −f(x)^3 ]^(1/3) for all x∈R. Then a period of f(x) is ka where k is a positive integer whose value is (A)1 (B)2 (C)3 (D)4

$$\mathrm{Let}\:{a}>\mathrm{0}\:\mathrm{and}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{a}\:\mathrm{function}\:\mathrm{satisfying} \\ $$$$\:\:\:\:\:{f}\left(\mathrm{x}+{a}\right)=\mathrm{1}+\left[\mathrm{2}−\mathrm{3}{f}\left(\mathrm{x}\right)+\mathrm{3}{f}\left(\mathrm{x}\right)^{\mathrm{2}} −{f}\left(\mathrm{x}\right)^{\mathrm{3}} \right]^{\mathrm{1}/\mathrm{3}} \\ $$$$\mathrm{for}\:\mathrm{all}\:\mathrm{x}\in\mathbb{R}.\:\mathrm{Then}\:\mathrm{a}\:\mathrm{period}\:\mathrm{of}\:{f}\left(\mathrm{x}\right)\:\mathrm{is}\:{ka}\:\mathrm{where}\:{k}\:\mathrm{is} \\ $$$$\mathrm{a}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{whose}\:\mathrm{value}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\mathrm{4} \\ $$

Question Number 119396    Answers: 3   Comments: 1

If the roots of the equation 24x^4 −52x^3 +18x^2 +13x−6=0 are α , −α , β and (1/β). Find the value of α and β.

$${If}\:{the}\:{roots}\:{of}\:{the}\:{equation}\: \\ $$$$\mathrm{24}{x}^{\mathrm{4}} −\mathrm{52}{x}^{\mathrm{3}} +\mathrm{18}{x}^{\mathrm{2}} +\mathrm{13}{x}−\mathrm{6}=\mathrm{0}\:{are}\: \\ $$$$\alpha\:,\:−\alpha\:,\:\beta\:{and}\:\frac{\mathrm{1}}{\beta}.\:{Find}\:{the}\:{value}\:{of}\: \\ $$$$\alpha\:{and}\:\beta. \\ $$

Question Number 119303    Answers: 3   Comments: 0

let x,y,z be positive real numbers such that x+y+z=1. Determine the minimum value of (1/x)+(4/y)+(9/z).

$$\:{let}\:{x},{y},{z}\:{be}\:{positive}\:{real}\:{numbers}\: \\ $$$${such}\:{that}\:{x}+{y}+{z}=\mathrm{1}.\:{Determine}\: \\ $$$${the}\:{minimum}\:{value}\:{of}\:\frac{\mathrm{1}}{{x}}+\frac{\mathrm{4}}{{y}}+\frac{\mathrm{9}}{{z}}. \\ $$

Question Number 119204    Answers: 0   Comments: 0

40−misolning yechimi: y=(x^3 /3)+2x^2 −5x+7 Kritik nuqtalarini topish uchun: 1. Funksiyadan hosila olamiz 2. Funksiya hosilasini nolga tenglab, tenglamani yechamiz. y′=x^2 +4x−5=0 ⇒ x_1 =1; x_2 =−5 x_1 +x_2 =1−5=−4 Javob: −4

$$\mathrm{40}−\mathrm{misolning}\:\:\:\mathrm{yechimi}: \\ $$$$\mathrm{y}=\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{2x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{7} \\ $$$$\mathrm{Kritik}\:\mathrm{nuqtalarini}\:\mathrm{topish}\:\mathrm{uchun}:\: \\ $$$$\mathrm{1}.\:\mathrm{Funksiyadan}\:\mathrm{hosila}\:\mathrm{olamiz} \\ $$$$\mathrm{2}.\:\mathrm{Funksiya}\:\mathrm{hosilasini}\:\mathrm{nolga}\:\mathrm{tenglab},\: \\ $$$$\mathrm{tenglamani}\:\mathrm{yechamiz}. \\ $$$$\mathrm{y}'=\mathrm{x}^{\mathrm{2}} +\mathrm{4x}−\mathrm{5}=\mathrm{0}\:\:\:\Rightarrow\:\:\:\mathrm{x}_{\mathrm{1}} =\mathrm{1};\:\:\mathrm{x}_{\mathrm{2}} =−\mathrm{5} \\ $$$$\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} =\mathrm{1}−\mathrm{5}=−\mathrm{4}\:\:\:\:\:\mathrm{Javob}:\:\:−\mathrm{4}\:\:\:\: \\ $$

Question Number 119160    Answers: 0   Comments: 5

Question Number 119159    Answers: 2   Comments: 0

We are in C. Given Z_(0 ) =1 ; Z_(n+1 ) =(1/2)Z_(n ) +(1/2)i n ∈ N. Show that ∀ n ∈ N^(∗ ) , ∣Z_n ∣<1.

$${We}\:{are}\:{in}\:\mathbb{C}. \\ $$$${Given}\:{Z}_{\mathrm{0}\:\:} =\mathrm{1}\:;\:\:\:\:\:{Z}_{{n}+\mathrm{1}\:} =\frac{\mathrm{1}}{\mathrm{2}}{Z}_{{n}\:} +\frac{\mathrm{1}}{\mathrm{2}}{i} \\ $$$${n}\:\in\:\mathbb{N}. \\ $$$${Show}\:{that}\:\forall\:{n}\:\in\:\mathbb{N}^{\ast\:} ,\:\mid{Z}_{{n}} \mid<\mathrm{1}. \\ $$

Question Number 119147    Answers: 0   Comments: 0

Question Number 119141    Answers: 1   Comments: 0

Question Number 119177    Answers: 1   Comments: 0

Question Number 119133    Answers: 0   Comments: 0

Informatica (11110000)_2 =0•2^0 +0•2^1 +0•2^2 +0•2^3 +1•2^4 +1•2^5 +1•2^6 +1•2^7 = =0•1+0•2+0•4+0•8+1•16+1•32+1•64+1•128= 0+0+0+0+16+32+64+128=(240)_(10) (11000101)_2 =1•2^0 +0•2^1 +1•2^2 +0•2^3 +0•2^4 +0•2^5 +1•2^6 +1•2^7 = = 1•1+0•2+1•4+0•8+0•16+0•32+1•64+1•128= (197)_(10) (01101001)_2 =

$$\mathrm{Informatica} \\ $$$$\left(\mathrm{11110000}\right)_{\mathrm{2}} =\mathrm{0}\bullet\mathrm{2}^{\mathrm{0}} +\mathrm{0}\bullet\mathrm{2}^{\mathrm{1}} +\mathrm{0}\bullet\mathrm{2}^{\mathrm{2}} +\mathrm{0}\bullet\mathrm{2}^{\mathrm{3}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{4}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{5}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{6}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{7}} = \\ $$$$=\mathrm{0}\bullet\mathrm{1}+\mathrm{0}\bullet\mathrm{2}+\mathrm{0}\bullet\mathrm{4}+\mathrm{0}\bullet\mathrm{8}+\mathrm{1}\bullet\mathrm{16}+\mathrm{1}\bullet\mathrm{32}+\mathrm{1}\bullet\mathrm{64}+\mathrm{1}\bullet\mathrm{128}= \\ $$$$\mathrm{0}+\mathrm{0}+\mathrm{0}+\mathrm{0}+\mathrm{16}+\mathrm{32}+\mathrm{64}+\mathrm{128}=\left(\mathrm{240}\right)_{\mathrm{10}} \\ $$$$\left(\mathrm{11000101}\right)_{\mathrm{2}} =\mathrm{1}\bullet\mathrm{2}^{\mathrm{0}} +\mathrm{0}\bullet\mathrm{2}^{\mathrm{1}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{2}} +\mathrm{0}\bullet\mathrm{2}^{\mathrm{3}} +\mathrm{0}\bullet\mathrm{2}^{\mathrm{4}} +\mathrm{0}\bullet\mathrm{2}^{\mathrm{5}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{6}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{7}} = \\ $$$$=\:\mathrm{1}\bullet\mathrm{1}+\mathrm{0}\bullet\mathrm{2}+\mathrm{1}\bullet\mathrm{4}+\mathrm{0}\bullet\mathrm{8}+\mathrm{0}\bullet\mathrm{16}+\mathrm{0}\bullet\mathrm{32}+\mathrm{1}\bullet\mathrm{64}+\mathrm{1}\bullet\mathrm{128}=\:\left(\mathrm{197}\right)_{\mathrm{10}} \\ $$$$\left(\mathrm{01101001}\right)_{\mathrm{2}} = \\ $$

Question Number 119119    Answers: 2   Comments: 0

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 )))

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\left(\frac{{x}!}{{x}^{{x}} }\right)^{\frac{\mathrm{1}}{{x}}} \right)^{\frac{\mathrm{2}\centerdot\mathrm{4}}{\mathrm{1}\centerdot\mathrm{3}}\centerdot\frac{\mathrm{4}\centerdot\mathrm{6}}{\mathrm{3}\centerdot\mathrm{5}}\centerdot\centerdot\centerdot} =? \\ $$$$\frac{\mathrm{2}}{\mathrm{1}}\centerdot\frac{\mathrm{2}}{\mathrm{3}}\centerdot\frac{\mathrm{4}}{\mathrm{3}}\centerdot\frac{\mathrm{4}}{\mathrm{5}}\centerdot\frac{\mathrm{6}}{\mathrm{5}}\centerdot\frac{\mathrm{6}}{\mathrm{7}}\centerdot\centerdot\centerdot=\frac{\pi}{\mathrm{2}} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\left(\frac{{x}!}{{x}^{{x}} }\right)^{\frac{\mathrm{1}}{{x}}} \right)^{\frac{\mathrm{2}}{\mathrm{1}}\centerdot\frac{\mathrm{4}}{\mathrm{3}}\centerdot\frac{\mathrm{4}}{\mathrm{3}}\centerdot\frac{\mathrm{6}}{\mathrm{5}}\centerdot\centerdot\centerdot} =\left(\frac{\mathrm{1}}{{e}}\right)^{\frac{\pi}{\mathrm{2}}} =\sqrt{\left(\frac{\mathrm{1}}{{e}}\right)^{\pi} }=\frac{\mathrm{1}}{\:\sqrt{{e}^{\pi} }} \\ $$

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.

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

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