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Question Number 131795    Answers: 1   Comments: 0

(1/1^3 )+(1/2^3 )+(1/5^3 )+(1/(10^3 ))+(1/(17^3 ))+(1/(26^3 ))+(1/(37^3 ))+(1/(50^3 ))+(1/(65^3 ))+(1/(82^3 ))+(1/(101^3 ))+...

$$\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{10}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{17}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{26}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{37}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{50}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{65}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{82}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{101}^{\mathrm{3}} }+... \\ $$

Question Number 131686    Answers: 1   Comments: 6

Σ_(n=1) ^∞ ((coth(nπ))/n^3 )

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{coth}\left({n}\pi\right)}{{n}^{\mathrm{3}} } \\ $$

Question Number 131580    Answers: 0   Comments: 2

Prove or disprove Σ_(n=0) ^∞ (1/((n^2 +97)^2 ))=(𝛑^2 /(97(e^(𝛑(√(97))) −e^(−𝛑(√(97))) )^2 ))+(𝛑/(388)).((e^(2𝛑(√(97))) +1)/(e^(2𝛑(√(97))) −1))+((37635)/(37636))−(1/( 388(√(97))))

$$\boldsymbol{\mathrm{Prove}}\:\boldsymbol{\mathrm{or}}\:\boldsymbol{\mathrm{disprove}} \\ $$$$\underset{\boldsymbol{{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\boldsymbol{{n}}^{\mathrm{2}} +\mathrm{97}\right)^{\mathrm{2}} }=\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{97}\left(\boldsymbol{{e}}^{\boldsymbol{\pi}\sqrt{\mathrm{97}}} −{e}^{−\boldsymbol{\pi}\sqrt{\mathrm{97}}} \right)^{\mathrm{2}} }+\frac{\boldsymbol{\pi}}{\mathrm{388}}.\frac{{e}^{\mathrm{2}\boldsymbol{\pi}\sqrt{\mathrm{97}}} +\mathrm{1}}{\boldsymbol{{e}}^{\mathrm{2}\boldsymbol{\pi}\sqrt{\mathrm{97}}} −\mathrm{1}}+\frac{\mathrm{37635}}{\mathrm{37636}}−\frac{\mathrm{1}}{\:\mathrm{388}\sqrt{\mathrm{97}}} \\ $$

Question Number 131507    Answers: 1   Comments: 0

1/Show that (2019)^(2021) +(2021)^(2019) divided by 2020 2/Show that 2222^(5555) +5555^(2222) divided by 7

$$ \\ $$$$\:\:\:\mathrm{1}/\mathrm{Show}\:\mathrm{that}\:\left(\mathrm{2019}\right)^{\mathrm{2021}} +\left(\mathrm{2021}\right)^{\mathrm{2019}} \:\mathrm{divided}\:\:\mathrm{by}\:\mathrm{2020} \\ $$$$\:\:\:\mathrm{2}/\mathrm{Show}\:\mathrm{that}\:\mathrm{2222}^{\mathrm{5555}} +\mathrm{5555}^{\mathrm{2222}} \:\mathrm{divided}\:\:\mathrm{by}\:\mathrm{7} \\ $$$$ \\ $$

Question Number 131346    Answers: 0   Comments: 0

Question Number 131094    Answers: 1   Comments: 1

Calculate 1/ I = ∮_c^+ ((zdz)/((z−1)^2 (z^2 −2z+1−2i))) ,C={z/∣z∣=2} 2/ J =∮_c^+ ((ch(z)dz)/(z(e^z −1))) , C={z/∣z−3i∣=4} 3/ K=∮_c^+ ((sin(z)dz)/(z^3 (z+1)^2 )) , C={z/∣z∣=2}

$$\:\: \\ $$$$\:\:\:\mathrm{Calculate} \\ $$$$\:\:\mathrm{1}/\:\mathrm{I}\:=\:\oint_{\mathrm{c}^{+} } \frac{\mathrm{zdz}}{\left(\mathrm{z}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{z}^{\mathrm{2}} −\mathrm{2z}+\mathrm{1}−\mathrm{2i}\right)}\:\:,\mathrm{C}=\left\{\mathrm{z}/\mid\mathrm{z}\mid=\mathrm{2}\right\}\: \\ $$$$\:\:\mathrm{2}/\:\mathrm{J}\:=\oint_{\mathrm{c}^{+} } \frac{\mathrm{ch}\left(\mathrm{z}\right)\mathrm{dz}}{\mathrm{z}\left(\mathrm{e}^{\mathrm{z}} −\mathrm{1}\right)}\:\:,\:\:\mathrm{C}=\left\{\mathrm{z}/\mid\mathrm{z}−\mathrm{3i}\mid=\mathrm{4}\right\} \\ $$$$\:\:\mathrm{3}/\:\mathrm{K}=\oint_{\mathrm{c}^{+} } \frac{\mathrm{sin}\left(\mathrm{z}\right)\mathrm{dz}}{\mathrm{z}^{\mathrm{3}} \left(\mathrm{z}+\mathrm{1}\right)^{\mathrm{2}} }\:\:,\:\mathrm{C}=\left\{\mathrm{z}/\mid\mathrm{z}\mid=\mathrm{2}\right\} \\ $$

Question Number 131083    Answers: 0   Comments: 0

1−(1/(5−((16)/(13−((81)/(25−((256)/(41−((625)/(61−((1296)/(85−((2401)/(113−...))))))))))))))=(6/𝛑^2 )

$$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}−\frac{\mathrm{16}}{\mathrm{13}−\frac{\mathrm{81}}{\mathrm{25}−\frac{\mathrm{256}}{\mathrm{41}−\frac{\mathrm{625}}{\mathrm{61}−\frac{\mathrm{1296}}{\mathrm{85}−\frac{\mathrm{2401}}{\mathrm{113}−...}}}}}}}=\frac{\mathrm{6}}{\boldsymbol{\pi}^{\mathrm{2}} } \\ $$

Question Number 130992    Answers: 1   Comments: 0

If I = (V/R) and V=250 volts and R=50 ohms Find the change in I resulting from an increase of 1 volt in V and increase of 0.5 ohm in R.

$$\mathrm{If}\:\mathrm{I}\:=\:\frac{\mathrm{V}}{\mathrm{R}}\:\mathrm{and}\:\mathrm{V}=\mathrm{250}\:\mathrm{volts}\:\mathrm{and}\:\mathrm{R}=\mathrm{50}\:\mathrm{ohms} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{change}\:\mathrm{in}\:\mathrm{I}\:\mathrm{resulting}\:\mathrm{from}\:\mathrm{an}\: \\ $$$$\mathrm{increase}\:\mathrm{of}\:\mathrm{1}\:\mathrm{volt}\:\mathrm{in}\:\mathrm{V}\:\mathrm{and}\:\mathrm{increase}\:\mathrm{of}\:\mathrm{0}.\mathrm{5}\:\mathrm{ohm} \\ $$$$\mathrm{in}\:\mathrm{R}. \\ $$

Question Number 131078    Answers: 0   Comments: 0

Σ_(n=1) ^∞ (1/(n^8 +1))

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{8}} +\mathrm{1}} \\ $$

Question Number 130978    Answers: 1   Comments: 5

Question Number 130925    Answers: 2   Comments: 4

Problem Without L′Hopital calculate lim_(x→0) ((tan^2 (x)−x^2 cos(2x))/(x^2 −sin^2 (x)))

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Problem} \\ $$$$\:\:\:\mathrm{Without}\:\:\mathrm{L}'\mathrm{Hopital} \\ $$$$\:\:\:\mathrm{calculate} \\ $$$$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tan}^{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{x}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{x}^{\mathrm{2}} −\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)} \\ $$

Question Number 130912    Answers: 0   Comments: 0

Σ_(n=1) ^∞ ((sech(nπ))/n)

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sech}\left({n}\pi\right)}{{n}} \\ $$

Question Number 130849    Answers: 3   Comments: 1

Question Number 130754    Answers: 0   Comments: 0

Question Number 130752    Answers: 0   Comments: 0

Question Number 130748    Answers: 1   Comments: 0

(1/3^3 )−(1/5^3 )+(1/(11^3 ))−(1/(13^3 ))+(1/(19^3 ))−(1/(21^3 ))+(1/(29^3 ))−(1/(31^3 ))+..

$$\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{11}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{13}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{19}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{21}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{29}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{31}^{\mathrm{3}} }+.. \\ $$

Question Number 130604    Answers: 3   Comments: 0

(d/dx)(x!)=???

$$\frac{{d}}{{dx}}\left({x}!\right)=??? \\ $$

Question Number 130555    Answers: 1   Comments: 0

(1/((2−π)^2 ))+(1/((2+π)^2 ))+(1/((6−π)^2 ))+(1/((6+π)^2 ))+(1/((10−π)^2 ))+(1/((10+π)^2 ))+..=(π^2 /(16))sec^2 ((π^2 /4)) Prove or disprove

$$\frac{\mathrm{1}}{\left(\mathrm{2}−\pi\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{2}+\pi\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{6}−\pi\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{6}+\pi\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{10}−\pi\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{10}+\pi\right)^{\mathrm{2}} }+..=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}{sec}^{\mathrm{2}} \left(\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$${Prove}\:{or}\:{disprove} \\ $$

Question Number 130549    Answers: 2   Comments: 0

the solution of equation ∣z∣−z = 1+2i is __

$${the}\:{solution}\:{of}\:{equation}\: \\ $$$$\mid{z}\mid−{z}\:=\:\mathrm{1}+\mathrm{2}{i}\:{is}\:\_\_ \\ $$

Question Number 130449    Answers: 0   Comments: 0

(1/((2−(√3))^3 ))−(1/((2+(√3))^3 ))+(1/((6−(√3))^3 ))−(1/((6+(√3))^3 ))+(1/((10−(√3))^3 ))−(1/((10+(√3))^3 ))+..

$$\frac{\mathrm{1}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }+\frac{\mathrm{1}}{\left(\mathrm{6}−\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left(\mathrm{6}+\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }+\frac{\mathrm{1}}{\left(\mathrm{10}−\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left(\mathrm{10}+\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }+.. \\ $$

Question Number 130415    Answers: 0   Comments: 1

(π/e)sin(1)−(π^2 /(2e^2 ))sin(2)+(π^3 /(3e^3 ))sin(3)−(π^4 /(4e^4 ))sin(4)+...

$$\frac{\pi}{{e}}{sin}\left(\mathrm{1}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{2}{e}^{\mathrm{2}} }{sin}\left(\mathrm{2}\right)+\frac{\pi^{\mathrm{3}} }{\mathrm{3}{e}^{\mathrm{3}} }{sin}\left(\mathrm{3}\right)−\frac{\pi^{\mathrm{4}} }{\mathrm{4}{e}^{\mathrm{4}} }{sin}\left(\mathrm{4}\right)+... \\ $$

Question Number 130399    Answers: 1   Comments: 0

Question Number 130369    Answers: 1   Comments: 0

Question Number 130370    Answers: 2   Comments: 0

Question Number 130337    Answers: 2   Comments: 0

(x^2 −4x+3)^(x^2 −6x+4) ≤ 1

$$\:\left(\mathrm{x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{3}\right)^{\mathrm{x}^{\mathrm{2}} −\mathrm{6x}+\mathrm{4}} \:\leqslant\:\mathrm{1}\: \\ $$

Question Number 130280    Answers: 0   Comments: 0

((d(uϕ))/dt)=?

$$\frac{{d}\left({u}\varphi\right)}{{dt}}=? \\ $$$$ \\ $$

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