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Question Number 197880 Answers: 1 Comments: 0

find the sum of infinite series (1/2^1 )∙(1/3^2 ) + (1/2^2 )∙(1/3^4 )(1^2 +2^2 +3^2 ) + (1/2^3 )∙(1/3^6 )(1^2 +2^2 +3^2 +...+7^2 )+ (1/2^4 )∙(1/3^8 )(1^2 +2^2 +3^2 +...+15^2 )+........

$$\mathrm{find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{infinite}\:\mathrm{series} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}} }\centerdot\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\centerdot\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{4}} }\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} \right)\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\centerdot\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{6}} }\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +...+\mathrm{7}^{\mathrm{2}} \right)+ \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }\centerdot\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{8}} }\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +...+\mathrm{15}^{\mathrm{2}} \right)+........ \\ $$

Question Number 197876 Answers: 2 Comments: 1

Question Number 197874 Answers: 1 Comments: 0

Question Number 197865 Answers: 1 Comments: 0

Question Number 197860 Answers: 3 Comments: 0

((2^(17) +2^(16) +2^(15) +…+1)/(2^8 +2^7 +2^6 +…+1)) = ?

$$\:\:\:\:\frac{\mathrm{2}^{\mathrm{17}} +\mathrm{2}^{\mathrm{16}} +\mathrm{2}^{\mathrm{15}} +\ldots+\mathrm{1}}{\mathrm{2}^{\mathrm{8}} +\mathrm{2}^{\mathrm{7}} +\mathrm{2}^{\mathrm{6}} +\ldots+\mathrm{1}}\:=\:?\: \\ $$

Question Number 197851 Answers: 0 Comments: 0

If z_1 = (√((α−β)/2)) and z_2 = (√((α+β)/2)) . Show that ∣z_1 −z_2 ∣^2 + ∣z_1 +z_2 ∣ = 2∣z∣^2 +∣z_2 ∣^2 and deduce that ∣α+(√(α^2 −β^2 ))∣ + ∣α−(√(α^2 −β^2 ))∣ = ∣α+β∣ + ∣α−β∣ Thank you in advance

$$\mathrm{If}\:\mathrm{z}_{\mathrm{1}} \:=\:\sqrt{\frac{\alpha−\beta}{\mathrm{2}}}\:\mathrm{and}\:\mathrm{z}_{\mathrm{2}} \:=\:\sqrt{\frac{\alpha+\beta}{\mathrm{2}}}\:.\:\mathrm{Show} \\ $$$$\mathrm{that}\:\mid\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:+\:\mid\mathrm{z}_{\mathrm{1}} +\mathrm{z}_{\mathrm{2}} \mid\:=\:\mathrm{2}\mid\mathrm{z}\mid^{\mathrm{2}} +\mid\mathrm{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:\mathrm{and} \\ $$$$\mathrm{deduce}\:\mathrm{that}\: \\ $$$$\mid\alpha+\sqrt{\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} }\mid\:+\:\mid\alpha−\sqrt{\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} }\mid\:=\:\mid\alpha+\beta\mid\:+\:\mid\alpha−\beta\mid \\ $$$$ \\ $$$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{in}\:\mathrm{advance} \\ $$

Question Number 197850 Answers: 1 Comments: 0

Solve the equation: (2sin x − 1)(2cos 2x + 2sin x +1) = 3(1−2cos 2x)

$${Solve}\:{the}\:{equation}: \\ $$$$\left(\mathrm{2}{sin}\:{x}\:−\:\mathrm{1}\right)\left(\mathrm{2}{cos}\:\mathrm{2}{x}\:+\:\mathrm{2}{sin}\:{x}\:+\mathrm{1}\right)\:=\:\mathrm{3}\left(\mathrm{1}−\mathrm{2}{cos}\:\mathrm{2}{x}\right) \\ $$

Question Number 197848 Answers: 4 Comments: 0

Question Number 197847 Answers: 2 Comments: 0

If (x−1)^2 +(y−(√3))^2 <1, then find the range of ((x+y)/( (√(x^2 +y^2 )))).

$$\mathrm{If}\:\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} <\mathrm{1},\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of} \\ $$$$\frac{{x}+{y}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}. \\ $$

Question Number 197843 Answers: 0 Comments: 2

Exercice 2

$$\boldsymbol{\mathrm{Exercice}}\:\:\mathrm{2} \\ $$

Question Number 197838 Answers: 2 Comments: 0

Question Number 197832 Answers: 1 Comments: 0

lim_(x→∞) ((2+(√(cosx)))/(−1+(√(cosx))))=?

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{2}+\sqrt{{cosx}}}{−\mathrm{1}+\sqrt{{cosx}}}=? \\ $$

Question Number 197834 Answers: 2 Comments: 0

Question Number 197822 Answers: 1 Comments: 2

find maximum of ∣z^2 +2z−3∣ ?

$${find}\:{maximum}\:{of}\:\mid{z}^{\mathrm{2}} +\mathrm{2}{z}−\mathrm{3}\mid\:? \\ $$

Question Number 197821 Answers: 1 Comments: 0

find the value of : Ω = ∫_0 ^( 1) (( ln ( 1+ (1/x^( 2) ) ))/(2 + x^( 2) )) dx = ?

$$ \\ $$$$\:\:\:\:{find}\:{the}\:{value}\:\:{of}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\:\mathrm{ln}\:\left(\:\mathrm{1}+\:\frac{\mathrm{1}}{{x}^{\:\mathrm{2}} }\:\right)}{\mathrm{2}\:+\:{x}^{\:\mathrm{2}} }\:{dx}\:=\:? \\ $$$$ \\ $$

Question Number 197819 Answers: 1 Comments: 0

find the value of : 𝛗 = Σ_(n=1) ^∞ (( (−1)^(n−1) H_( 2n) )/n) = ? where,H_n =1+(1/2) +(1/3) +...+(1/n)

$$ \\ $$$$\:\:\:\:\:\:{find}\:{the}\:{value}\:{of}\:\:: \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:{H}_{\:\mathrm{2}{n}} }{{n}}\:=\:? \\ $$$${where},{H}_{{n}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}\:+...+\frac{\mathrm{1}}{{n}} \\ $$

Question Number 197808 Answers: 1 Comments: 1

prove lim_(n→∞) x^n = 0 when ∣x∣ < 1

$${prove}\:\underset{{n}\rightarrow\infty} {{lim}}\:{x}^{{n}} \:=\:\mathrm{0}\:\:\:\:{when}\:\mid{x}\mid\:<\:\mathrm{1} \\ $$

Question Number 197802 Answers: 1 Comments: 0

I=∫_(−2) ^6 ((∣x−1∣)/(x−1)) dx =?

$$\:\:\:\mathrm{I}=\underset{−\mathrm{2}} {\overset{\mathrm{6}} {\int}}\:\frac{\mid\mathrm{x}−\mathrm{1}\mid}{\mathrm{x}−\mathrm{1}}\:\mathrm{dx}\:=? \\ $$

Question Number 197906 Answers: 1 Comments: 0

S= Σ_(k=1) ^∞ (( Γ^( 2) ( k ))/(k Γ (2k ))) = ? −−−−

$$ \\ $$$$\:\:\:\:\:\:\:\mathrm{S}=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\:\Gamma^{\:\mathrm{2}} \left(\:{k}\:\right)}{{k}\:\Gamma\:\left(\mathrm{2}{k}\:\right)}\:=\:? \\ $$$$\:\:\:\:\:\:\:\:−−−− \\ $$

Question Number 197795 Answers: 2 Comments: 0

Question Number 197794 Answers: 1 Comments: 0

if x = log tan((π/4)+(y/2)), prove that y = −ilog tan(((ix)/2) + (π/4)) here i = (√(−1))

$$\:\:\:\:\mathrm{if}\:\mathrm{x}\:\:\:=\:\:\:\mathrm{log}\:\mathrm{tan}\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{y}}{\mathrm{2}}\right),\:\:\mathrm{prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\mathrm{y}\:\:\:\:=\:\:\:−{i}\mathrm{log}\:\mathrm{tan}\left(\frac{{ix}}{\mathrm{2}}\:+\:\frac{\pi}{\mathrm{4}}\right)\:\:\:\:\:\mathrm{here}\:{i}\:\:=\:\sqrt{−\mathrm{1}} \\ $$

Question Number 197792 Answers: 2 Comments: 0

Solve the following differential equation 1) y′′ + y = e^x + x^3 , y(0)=2, y′(0)=0 2) y′′ + y^′ − 2y = x + sin2x, y(0)=1, y′(0)=0 3) y′′ − y′ = xe^x , y(0)=2, y′(0)= 1 Thank you

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{following}\:\mathrm{differential}\:\mathrm{equation} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{y}''\:+\:\mathrm{y}\:=\:\mathrm{e}^{\mathrm{x}} \:+\:\mathrm{x}^{\mathrm{3}} ,\:\:\:\:\:\:\:\:\:\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{2},\:\mathrm{y}'\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{y}''\:+\:\mathrm{y}^{'} \:−\:\mathrm{2y}\:=\:\mathrm{x}\:+\:\mathrm{sin2x},\:\:\:\:\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{1},\:\mathrm{y}'\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\:\mathrm{y}''\:−\:\mathrm{y}'\:=\:\mathrm{xe}^{\mathrm{x}} ,\:\:\:\:\:\:\:\:\:\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{2},\:\mathrm{y}'\left(\mathrm{0}\right)=\:\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Thank}\:\mathrm{you} \\ $$

Question Number 197784 Answers: 1 Comments: 0

((x−2+3((x−3))^(1/3) (1+((x−3))^(1/3) )))^(1/3) + ((x+5+6((x−3))^(1/3) (1+2((x−3))^(1/3) )))^(1/3) = 5

$$\:\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{2}+\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{3}}\:\left(\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{3}}\right)}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{x}+\mathrm{5}+\mathrm{6}\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{3}}\left(\mathrm{1}+\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{3}}\:\right)}\:=\:\mathrm{5} \\ $$

Question Number 197783 Answers: 2 Comments: 0

∫((x.arctg(x))/(x^2 +1))dx=?

$$\int\frac{{x}.\boldsymbol{{arctg}}\left(\boldsymbol{{x}}\right)}{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{1}}\boldsymbol{{dx}}=? \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 197776 Answers: 1 Comments: 3

Question Number 197772 Answers: 2 Comments: 1

Can anyone do this? ∫^( +∞) _( 1) ((t−1)/((1+t)^3 lnt))dt

$$\mathrm{Can}\:\mathrm{anyone}\:\mathrm{do}\:\mathrm{this}? \\ $$$$\underset{\:\mathrm{1}} {\int}^{\:+\infty} \frac{\mathrm{t}−\mathrm{1}}{\left(\mathrm{1}+\mathrm{t}\right)^{\mathrm{3}} \:\mathrm{lnt}}\mathrm{dt} \\ $$

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