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Question Number 127859    Answers: 1   Comments: 0

Question Number 127793    Answers: 2   Comments: 1

Σ_(n=1) ^∞ (n/(n!))cos(((πn)/5))

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}}{{n}!}{cos}\left(\frac{\pi{n}}{\mathrm{5}}\right) \\ $$

Question Number 127771    Answers: 0   Comments: 3

(1/(1−(π^2 /(1+π^2 −((2π^2 )/(2+π^2 −((3π^2 )/(3+π^2 −((4π^2 )/(4+π^2 −((5π^2 )/(5+π^2 ....))))))))))))

$$\frac{\mathrm{1}}{\mathrm{1}−\frac{\pi^{\mathrm{2}} }{\mathrm{1}+\pi^{\mathrm{2}} −\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{2}+\pi^{\mathrm{2}} −\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{3}+\pi^{\mathrm{2}} −\frac{\mathrm{4}\pi^{\mathrm{2}} }{\mathrm{4}+\pi^{\mathrm{2}} −\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{5}+\pi^{\mathrm{2}} ....}}}}}} \\ $$

Question Number 127982    Answers: 0   Comments: 1

Some Values .. Σ_(n=−∞) ^∞ e^(−πn^2 ) =(π^(1/4) /(Γ((3/4)))) Σ_(n=−∞) ^∞ e^(−2πn^2 ) =(π^(1/4) /(Γ((3/4)))) (((6+4(√2)))^(1/4) /2) Σ_(n=−∞) ^∞ e^(−6πn^2 ) =(π^(1/4) /(Γ((3/4)))).((√((1)^(1/4) +(3)^(1/4) +(4)^(1/4) +(9)^(1/4) ))/( (√(1728)))) Any Idea to prove ?

$${Some}\:{Values}\:.. \\ $$$$\underset{{n}=−\infty} {\overset{\infty} {\sum}}{e}^{−\pi{n}^{\mathrm{2}} } =\frac{\pi^{\frac{\mathrm{1}}{\mathrm{4}}} }{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)} \\ $$$$\underset{{n}=−\infty} {\overset{\infty} {\sum}}{e}^{−\mathrm{2}\pi{n}^{\mathrm{2}} } =\frac{\pi^{\frac{\mathrm{1}}{\mathrm{4}}} }{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}\:\frac{\sqrt[{\mathrm{4}}]{\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$$\underset{{n}=−\infty} {\overset{\infty} {\sum}}{e}^{−\mathrm{6}\pi{n}^{\mathrm{2}} } =\frac{\pi^{\frac{\mathrm{1}}{\mathrm{4}}} }{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}.\frac{\sqrt{\sqrt[{\mathrm{4}}]{\mathrm{1}}+\sqrt[{\mathrm{4}}]{\mathrm{3}}+\sqrt[{\mathrm{4}}]{\mathrm{4}}+\sqrt[{\mathrm{4}}]{\mathrm{9}}}}{\:\sqrt{\mathrm{1728}}} \\ $$$${Any}\:{Idea}\:{to}\:{prove}\:? \\ $$

Question Number 127682    Answers: 1   Comments: 2

(π^2 /(1+(π^2 /(3−π^2 +((9π^2 )/(5−3π^2 +((25π^2 )/(7−5π^2 +((49π^( 2) )/(9−7π^2 +((81π^2 )/(11−9π^2 +((121π^2 )/(.....))))))))))))))

$$\frac{\pi^{\mathrm{2}} }{\mathrm{1}+\frac{\pi^{\mathrm{2}} }{\mathrm{3}−\pi^{\mathrm{2}} +\frac{\mathrm{9}\pi^{\mathrm{2}} }{\mathrm{5}−\mathrm{3}\pi^{\mathrm{2}} +\frac{\mathrm{25}\pi^{\mathrm{2}} }{\mathrm{7}−\mathrm{5}\pi^{\mathrm{2}} +\frac{\mathrm{49}\pi^{\:\mathrm{2}} }{\mathrm{9}−\mathrm{7}\pi^{\mathrm{2}} +\frac{\mathrm{81}\pi^{\mathrm{2}} }{\mathrm{11}−\mathrm{9}\pi^{\mathrm{2}} +\frac{\mathrm{121}\pi^{\mathrm{2}} }{.....}}}}}}} \\ $$

Question Number 127666    Answers: 0   Comments: 11

Have a great year all of you It is 12.20 am in India (GMT+5.30) (12.20)^T =20.21 💐🌅

$${Have}\:{a}\:{great}\:{year}\:{all}\:{of}\:{you} \\ $$$${It}\:{is}\:\mathrm{12}.\mathrm{20}\:{am}\:{in}\:{India}\:\left({GMT}+\mathrm{5}.\mathrm{30}\right) \\ $$$$\left(\mathrm{12}.\mathrm{20}\right)^{{T}} =\mathrm{20}.\mathrm{21} \\ $$$$ \\ $$💐🌅

Question Number 127644    Answers: 0   Comments: 2

(1/(1−(1/(2−((1/2)/((3/2)−((1/3)/((4/3)−((1/4)/((5/4)−((1/5)/((6/5)−...))))))))))))

$$\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}−\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\mathrm{3}}{\mathrm{2}}−\frac{\frac{\mathrm{1}}{\mathrm{3}}}{\frac{\mathrm{4}}{\mathrm{3}}−\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\frac{\mathrm{5}}{\mathrm{4}}−\frac{\frac{\mathrm{1}}{\mathrm{5}}}{\frac{\mathrm{6}}{\mathrm{5}}−...}}}}}} \\ $$

Question Number 127552    Answers: 0   Comments: 0

Question Number 127295    Answers: 0   Comments: 0

∫_0 ^(1/2) ((tanh^(−1) x)/( (x)^(1/5) ))dx

$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{tanh}^{−\mathrm{1}} {x}}{\:\sqrt[{\mathrm{5}}]{{x}}}{dx}\:\:\:\:\:\:\: \\ $$

Question Number 127293    Answers: 0   Comments: 0

Σ_(n=1) ^∞ ((coth(n))/n^3 )

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

Question Number 127187    Answers: 0   Comments: 1

1−5((1/2))^3 +9((1/2).(3/4))^3 −13((1/2).(3/4).(5/6))^3 +..=(2/π) (prove)

$$\mathrm{1}−\mathrm{5}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} +\mathrm{9}\left(\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{3}} −\mathrm{13}\left(\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{3}}{\mathrm{4}}.\frac{\mathrm{5}}{\mathrm{6}}\right)^{\mathrm{3}} +..=\frac{\mathrm{2}}{\pi}\:\left({prove}\right) \\ $$

Question Number 127186    Answers: 0   Comments: 1

∫_0 ^a e^(−x^2 ) dx=((√π)/2)−(e^(−a^2 ) /(2a+(1/(a+(2/(2a+(3/(a+(4/(2a+...)))))))))) (Prove)

$$\int_{\mathrm{0}} ^{{a}} {e}^{−{x}^{\mathrm{2}} } {dx}=\frac{\sqrt{\pi}}{\mathrm{2}}−\frac{{e}^{−{a}^{\mathrm{2}} } }{\mathrm{2}{a}+\frac{\mathrm{1}}{{a}+\frac{\mathrm{2}}{\mathrm{2}{a}+\frac{\mathrm{3}}{{a}+\frac{\mathrm{4}}{\mathrm{2}{a}+...}}}}}\:\left({Prove}\right) \\ $$

Question Number 127080    Answers: 0   Comments: 1

Σ_(n=1) ^∞ (1/(e^(−φn) +((e^(2πn) −e^(−2φn) )/(2e^(−φn) +((e^(2πn) −e^(−2φn) )/(2e^(−φn) +((e^(2πn) −e^(−2φn) )/(2e^(−2φn) ...))))))))

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{e}^{−\phi{n}} +\frac{{e}^{\mathrm{2}\pi{n}} −{e}^{−\mathrm{2}\phi{n}} \:}{\mathrm{2}{e}^{−\phi{n}} +\frac{{e}^{\mathrm{2}\pi{n}} −{e}^{−\mathrm{2}\phi{n}} }{\mathrm{2}{e}^{−\phi{n}} +\frac{{e}^{\mathrm{2}\pi{n}} −{e}^{−\mathrm{2}\phi{n}} }{\mathrm{2}{e}^{−\mathrm{2}\phi{n}} ...}}}} \\ $$

Question Number 127004    Answers: 0   Comments: 1

Question Number 126977    Answers: 1   Comments: 1

Σ_(n=1) ^∞ (n^7 /7^n )

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{7}} }{\mathrm{7}^{{n}} } \\ $$

Question Number 126934    Answers: 1   Comments: 3

(1/(1!))+((1!^2 )/(3!))+((2!^2 )/(5!))+((3!^2 )/(7!))+((4!^2 )/(9!))+....

$$\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}!^{\mathrm{2}} }{\mathrm{3}!}+\frac{\mathrm{2}!^{\mathrm{2}} }{\mathrm{5}!}+\frac{\mathrm{3}!^{\mathrm{2}} }{\mathrm{7}!}+\frac{\mathrm{4}!^{\mathrm{2}} }{\mathrm{9}!}+.... \\ $$

Question Number 126913    Answers: 1   Comments: 1

to Tinku tara equation editor is not available in playstore now...pls check..i suggested a few students to dowmload it

$${to}\:{Tinku}\:{tara} \\ $$$${equation}\:{editor}\:{is}\:{not}\:{available}\:{in}\:{playstore} \\ $$$${now}...{pls}\:{check}..{i}\:{suggested}\:{a}\:{few}\:{students} \\ $$$${to}\:{dowmload}\:{it} \\ $$

Question Number 126907    Answers: 1   Comments: 0

Merry christmas !! 🎅🤶☃️🌄🎄🦌 🔔🔔🔔🔔🔔🔔🔔🔔🔔 🎄🎄🎄🎄🎄🎄🎄🎄 ∫_0 ^(1/2) ((tanh^(−1) x)/( (x)^(1/5) ))dx

$$\boldsymbol{{Merry}}\:\boldsymbol{{christmas}}\:!! \\ $$$$ \\ $$🎅🤶☃️🌄🎄🦌 $$ \\ $$$$ \\ $$🔔🔔🔔🔔🔔🔔🔔🔔🔔 🎄🎄🎄🎄🎄🎄🎄🎄 $$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\boldsymbol{{tanh}}^{−\mathrm{1}} \boldsymbol{{x}}}{\:\sqrt[{\mathrm{5}}]{\boldsymbol{{x}}}}\boldsymbol{{dx}} \\ $$

Question Number 126780    Answers: 1   Comments: 1

Question Number 126704    Answers: 0   Comments: 0

((e^π −1)/(e^π +1))=(π/(2+(π^2 /(6+(π^2 /(10+(π^2 /(14+....))))))))

$$\frac{{e}^{\pi} −\mathrm{1}}{{e}^{\pi} +\mathrm{1}}=\frac{\pi}{\mathrm{2}+\frac{\pi^{\mathrm{2}} }{\mathrm{6}+\frac{\pi^{\mathrm{2}} }{\mathrm{10}+\frac{\pi^{\mathrm{2}} }{\mathrm{14}+....}}}} \\ $$

Question Number 192126    Answers: 2   Comments: 1

prove that ∣z∣ > ((∣Re(z)∣ +∣Im(z)∣)/2) , ∀z∈C

$$\:\:\:\:\boldsymbol{{prove}}\:\boldsymbol{{that}} \\ $$$$\:\:\:\:\:\:\mid\boldsymbol{{z}}\mid\:>\:\frac{\mid\boldsymbol{{Re}}\left(\boldsymbol{{z}}\right)\mid\:+\mid\boldsymbol{{Im}}\left(\boldsymbol{{z}}\right)\mid}{\mathrm{2}}\:\:,\:\:\:\forall\boldsymbol{{z}}\in\mathbb{C} \\ $$

Question Number 126669    Answers: 1   Comments: 2

Σ_(n=1) ^∞ (H_n ^2 /n^4 )

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{H}_{{n}} ^{\mathrm{2}} }{{n}^{\mathrm{4}} } \\ $$

Question Number 126632    Answers: 2   Comments: 1

Question Number 126400    Answers: 1   Comments: 5

If a_n =6^n +8^n find (a_(1991) /(49)).

$${If}\:{a}_{{n}} =\mathrm{6}^{{n}} +\mathrm{8}^{{n}} \:{find}\:\frac{{a}_{\mathrm{1991}} }{\mathrm{49}}. \\ $$

Question Number 126200    Answers: 0   Comments: 3

(1/1^2 )−(1/2^3 )+(1/3^5 )−(1/4^7 )+(1/5^(11) )−(1/6^(13) )+....

$$\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{5}} }−\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{7}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{11}} }−\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{13}} }+.... \\ $$

Question Number 126092    Answers: 2   Comments: 0

A particle starts from rest at time t = 0 and moves in a straightline with variable acceleration a m/s^2 where a = (t/5) , 0 ≤ t ≤ 5 , a = (t/5) + ((10)/t^2 ) , t ≥ 5, t being measured in seconds. Show that the velocity is 22(1/2) m/s when t = 5 and 11 m/s when t = 10. Show also that the distance travelled by the particle in the first 10 seconds is (43(1/3)−10 ln 2) m.

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{starts}\:\mathrm{from}\:\mathrm{rest}\:\mathrm{at}\:\mathrm{time}\:{t}\:=\:\mathrm{0}\:\mathrm{and}\:\mathrm{moves}\:\mathrm{in}\: \\ $$$$\mathrm{a}\:\mathrm{straightline}\:\mathrm{with}\:\mathrm{variable}\:\mathrm{acceleration}\:{a}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \:\mathrm{where}\: \\ $$$$\:{a}\:=\:\frac{{t}}{\mathrm{5}}\:,\:\mathrm{0}\:\leqslant\:{t}\:\leqslant\:\mathrm{5}\:,\:{a}\:=\:\frac{{t}}{\mathrm{5}}\:+\:\frac{\mathrm{10}}{{t}^{\mathrm{2}} }\:,\:{t}\:\geqslant\:\mathrm{5},\:{t}\:\mathrm{being}\:\mathrm{measured}\:\mathrm{in}\:\mathrm{seconds}. \\ $$$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{velocity}\:\mathrm{is}\:\mathrm{22}\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{m}/\mathrm{s}\:\mathrm{when}\:{t}\:=\:\mathrm{5}\:\mathrm{and} \\ $$$$\mathrm{11}\:\mathrm{m}/\mathrm{s}\:\mathrm{when}\:{t}\:=\:\mathrm{10}. \\ $$$$\mathrm{Show}\:\mathrm{also}\:\mathrm{that}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{travelled}\:\mathrm{by}\:\mathrm{the}\:\mathrm{particle} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{first}\:\mathrm{10}\:\mathrm{seconds}\:\mathrm{is}\:\:\left(\mathrm{43}\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{10}\:\mathrm{ln}\:\mathrm{2}\right)\:\mathrm{m}. \\ $$

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