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Question Number 194020    Answers: 0   Comments: 0

F_n = F_n _(−1) +F_(n−2) F_2 = F_1 =1 F_n : 1 , 1 , 2 , 3 ,5... f(x)= Σ_(n=1) ^∞ F_n x^( n) = x + x^( 2) +Σ_(n=3) ^∞ (F_(n−1) +F_(n−2) )x^( n) = x+x^2 + Σ_(n=3) ^∞ F_(n−1) x^( n) + x^( 2) f (x) = x + x^( 2) + x^( 2) f(x) +x Σ_(n=2) ^∞ F_n x^( n) = x + x^( 2) + x^( 2) f(x)−x^( 2) + xf(x) ∴ f(x)= (x/(1−x−x^( 2) )) (generating function ) (x/(1−x−x^( 2) )) =Σ_(n=1) ^∞ F_n x^( n) ⇒ (x^( 2) /(1−x−x^( 2) ))=Σ_(n=1) ^∞ F_n x^( n+1) x= (1/(10)) ⇒ ((1/(100))/(1−(1/(10))−(1/(100)))) = Σ_(n=1) ^∞ (F_n /(10^( n+1) )) ⇒ { Σ_(n=1) ^∞ (( F_n )/(10^( n+1) )) = (1/(89)) }

$$\:\:\:\:\:\:{F}_{{n}} =\:{F}_{{n}} \:_{−\mathrm{1}} +{F}_{{n}−\mathrm{2}} \:\:\:\:\:\:{F}_{\mathrm{2}} =\:{F}_{\mathrm{1}} =\mathrm{1}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:{F}_{{n}} \::\:\:\:\:\mathrm{1}\:,\:\mathrm{1}\:,\:\mathrm{2}\:,\:\mathrm{3}\:,\mathrm{5}...\:\: \\ $$$$\:\:\:\:\:\:\:{f}\left({x}\right)=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:{F}_{{n}} \:{x}^{\:{n}} \:=\:{x}\:+\:{x}^{\:\mathrm{2}} \:+\underset{{n}=\mathrm{3}} {\overset{\infty} {\sum}}\left({F}_{{n}−\mathrm{1}} +{F}_{{n}−\mathrm{2}} \right){x}^{\:{n}} \\ $$$$\:\:\:=\:\:{x}+{x}^{\mathrm{2}} \:+\:\underset{{n}=\mathrm{3}} {\overset{\infty} {\sum}}{F}_{{n}−\mathrm{1}} {x}^{\:{n}} \:+\:{x}^{\:\mathrm{2}} \:{f}\:\left({x}\right) \\ $$$$\:\:\:\:=\:{x}\:+\:{x}^{\:\mathrm{2}} \:+\:{x}^{\:\mathrm{2}} {f}\left({x}\right)\:+{x}\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}{F}_{{n}} \:{x}^{\:{n}} \\ $$$$\:\:\:=\:{x}\:+\:{x}^{\:\mathrm{2}} \:+\:{x}^{\:\mathrm{2}} {f}\left({x}\right)−{x}^{\:\mathrm{2}} +\:{xf}\left({x}\right) \\ $$$$ \\ $$$$\:\: \\ $$$$\:\: \\ $$$$\:\:\therefore\:\:\:{f}\left({x}\right)=\:\frac{{x}}{\mathrm{1}−{x}−{x}^{\:\mathrm{2}} }\:\:\:\left({generating}\:{function}\:\right)\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{{x}}{\mathrm{1}−{x}−{x}^{\:\mathrm{2}} }\:\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{F}_{{n}} {x}^{\:{n}} \:\Rightarrow\:\frac{{x}^{\:\mathrm{2}} }{\mathrm{1}−{x}−{x}^{\:\mathrm{2}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{F}_{{n}} \:{x}^{\:{n}+\mathrm{1}} \\ $$$$\:\:\:{x}=\:\frac{\mathrm{1}}{\mathrm{10}}\:\Rightarrow\:\:\frac{\frac{\mathrm{1}}{\mathrm{100}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{10}}−\frac{\mathrm{1}}{\mathrm{100}}}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{F}_{{n}} }{\mathrm{10}^{\:{n}+\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:\Rightarrow\:\:\left\{\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\:{F}_{{n}} }{\mathrm{10}^{\:{n}+\mathrm{1}} }\:=\:\frac{\mathrm{1}}{\mathrm{89}}\:\:\:\:\:\right\}\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\: \\ $$

Question Number 194017    Answers: 1   Comments: 0

Question Number 194015    Answers: 1   Comments: 0

Question Number 194011    Answers: 0   Comments: 0

Give formula of the following complexes Compound (a)Diammine Silver (i) Chloride (b)Dichloro aqua tetra ammine Cobalt(iii) Chloride (c)Tetra ammine (ii) hexachloro platinum (iv) (d)Bromo cyano fluoro Carbonly Iron (iii)

$${Give}\:{formula}\:{of}\:{the}\:{following}\:{complexes} \\ $$$${Compound} \\ $$$$\left({a}\right){Diammine}\:{Silver}\:\left({i}\right)\:{Chloride} \\ $$$$ \\ $$$$\left({b}\right){Dichloro}\:{aqua}\:{tetra}\:{ammine}\:{Cobalt}\left({iii}\right)\:{Chloride} \\ $$$$ \\ $$$$\left({c}\right){Tetra}\:{ammine}\:\left({ii}\right)\:{hexachloro}\:{platinum}\:\left({iv}\right) \\ $$$$ \\ $$$$\left({d}\right){Bromo}\:{cyano}\:{fluoro}\:{Carbonly}\:{Iron}\:\left({iii}\right) \\ $$

Question Number 193987    Answers: 1   Comments: 0

Question Number 193984    Answers: 0   Comments: 1

Know f(x^(−1) )=f^( −1) (x) Find f(x)¿

$${Know}\:{f}\left({x}^{−\mathrm{1}} \right)={f}^{\:−\mathrm{1}} \left({x}\right) \\ $$$${Find}\:{f}\left({x}\right)¿ \\ $$

Question Number 193982    Answers: 3   Comments: 0

∫ ((6x^3 +9x^2 +15x+6)/( (√(x^2 +x+1)))) dx =?

$$\:\:\:\:\:\int\:\frac{\mathrm{6x}^{\mathrm{3}} +\mathrm{9x}^{\mathrm{2}} +\mathrm{15x}+\mathrm{6}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}}\:\mathrm{dx}\:=? \\ $$

Question Number 193981    Answers: 0   Comments: 0

Question Number 193978    Answers: 1   Comments: 0

Question Number 193976    Answers: 2   Comments: 0

Question Number 193975    Answers: 2   Comments: 1

Question Number 193972    Answers: 0   Comments: 4

Question Number 193971    Answers: 2   Comments: 0

y= ⌊ x^( 2) ⌋ + ⌊ x ⌋ ⇒ R_y =? Range

$$ \\ $$$$\:\:\:\:\:\:{y}=\:\lfloor\:{x}^{\:\mathrm{2}} \:\rfloor\:+\:\lfloor\:{x}\:\rfloor\:\: \\ $$$$\:\: \\ $$$$\:\:\:\:\:\:\:\Rightarrow\:\:{R}_{{y}} \:=?\:\:\:\:\:\:\:{Range} \\ $$

Question Number 193965    Answers: 2   Comments: 0

a,b,c are positive real numbers And (1/(a+b+1))+(1/(b+c+1))+(1/(a+c+1))≥1 prove that a+b+c≥ab+bc+ac

$${a},{b},{c}\:{are}\:{positive}\:{real}\:{numbers} \\ $$$${And} \\ $$$$\frac{\mathrm{1}}{{a}+{b}+\mathrm{1}}+\frac{\mathrm{1}}{{b}+{c}+\mathrm{1}}+\frac{\mathrm{1}}{{a}+{c}+\mathrm{1}}\geqslant\mathrm{1} \\ $$$${prove}\:{that}\:{a}+{b}+{c}\geqslant{ab}+{bc}+{ac} \\ $$

Question Number 193962    Answers: 2   Comments: 0

Question Number 194691    Answers: 0   Comments: 0

a, b, c≥0, a+b+c=2. Prove that 3a+8ab+16abc≤12.

$${a},\:{b},\:{c}\geqslant\mathrm{0},\:{a}+{b}+{c}=\mathrm{2}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{3}{a}+\mathrm{8}{ab}+\mathrm{16}{abc}\leqslant\mathrm{12}. \\ $$

Question Number 194689    Answers: 1   Comments: 0

Question Number 194688    Answers: 1   Comments: 0

find all function f: R → R such that ∀x, y∈R, f(x−f(y))=f(f(y))+xf(y)+f(x)−1.

$$\mathrm{find}\:\mathrm{all}\:\mathrm{function}\:{f}:\:\mathbb{R}\:\rightarrow\:\mathbb{R}\:\mathrm{such}\:\mathrm{that}\:\forall{x},\:{y}\in\mathbb{R}, \\ $$$${f}\left({x}−{f}\left({y}\right)\right)={f}\left({f}\left({y}\right)\right)+{xf}\left({y}\right)+{f}\left({x}\right)−\mathrm{1}. \\ $$

Question Number 193958    Answers: 1   Comments: 0

Question Number 193951    Answers: 2   Comments: 0

Prove that: lim_(n→+∞) (1/(n!))∫^( n) _( 0) t^n e^(−t) dt = (1/2)

$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\underset{\mathrm{n}\rightarrow+\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{n}!}\underset{\:\mathrm{0}} {\int}^{\:\mathrm{n}} {t}^{{n}} {e}^{−{t}} {dt}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Question Number 193953    Answers: 3   Comments: 0

Name thefollowing Complex compound (a)[Zn(en)_2 (c_2 o_2 )Br_2 F]^(−5) (b)Ni(CN)_2 (OH)_3 Cl] [Fe(H_2 O)_3 (OH)_2 F_2 Br] (c)[Cr(CN)_4 NO_2 (H_2 O)_2 (NH_3 )_2 ]^(2−)

$${Name}\:{thefollowing}\:{Complex}\:{compound} \\ $$$$\left({a}\right)\left[{Zn}\left({en}\right)_{\mathrm{2}} \left({c}_{\mathrm{2}} {o}_{\mathrm{2}} \right){Br}_{\mathrm{2}} {F}\right]^{−\mathrm{5}} \\ $$$$\left.\left({b}\right){Ni}\left({CN}\right)_{\mathrm{2}} \left({OH}\right)_{\mathrm{3}} {Cl}\right]\:\:\left[{Fe}\left({H}_{\mathrm{2}} {O}\right)_{\mathrm{3}} \left({OH}\right)_{\mathrm{2}} {F}_{\mathrm{2}} {Br}\right] \\ $$$$\left({c}\right)\left[{Cr}\left({CN}\right)_{\mathrm{4}} {NO}_{\mathrm{2}} \left({H}_{\mathrm{2}} {O}\right)_{\mathrm{2}} \left({NH}_{\mathrm{3}} \right)_{\mathrm{2}} \:\right]^{\mathrm{2}−} \\ $$

Question Number 193949    Answers: 3   Comments: 0

Question Number 193938    Answers: 4   Comments: 0

Question Number 193924    Answers: 1   Comments: 2

question about tinkutara how can an answer be placed in a box.

$$\:\:\underline{\mathrm{question}\:\mathrm{about}\:\mathrm{tinkutara}} \\ $$$$\:\:\mathrm{how}\:\mathrm{can}\:\mathrm{an}\:\mathrm{answer}\:\mathrm{be}\:\mathrm{placed}\:\: \\ $$$$\:\:\mathrm{in}\:\mathrm{a}\:\mathrm{box}. \\ $$

Question Number 193922    Answers: 4   Comments: 0

Question Number 193921    Answers: 0   Comments: 0

Show that the kernel of a group homomorhism θ : G → H is a normal subgroup. Hint: Check the existence of the combination g^(−1) kg in the kernel.

$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{kernel}\:\mathrm{of}\:\mathrm{a}\:\mathrm{group}\:\mathrm{homomorhism} \\ $$$$\theta\::\:\mathrm{G}\:\rightarrow\:\mathrm{H}\:\mathrm{is}\:\mathrm{a}\:\mathrm{normal}\:\mathrm{subgroup}. \\ $$$$\mathrm{Hint}:\:\mathrm{Check}\:\mathrm{the}\:\mathrm{existence}\:\mathrm{of}\:\mathrm{the}\:\mathrm{combination} \\ $$$$\mathrm{g}^{−\mathrm{1}} \mathrm{kg}\:\mathrm{in}\:\mathrm{the}\:\mathrm{kernel}. \\ $$

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