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Question Number 111813    Answers: 0   Comments: 3

if ψ(x)=(x/(1−x))+(x^3 /(1−x^2 ))+(x^5 /(1−x^5 ))+(x^7 /(1−x^7 )) show that ∫_0 ^1 [ψ(x)+ψ((x)^(1/3) )+ψ((x)^(1/5) )+ψ((x)^(1/7) )]((ln((1/x)))/x)=((25832π^2 )/(1575)) soltion let the generating series form of ψ(x)=(x/(1−x))+(x^3 /(1−x^2 ))+(x^5 /(1−x^5 ))+(x^7 /(1−x^7 )) be ψ(x)=Σ_(i=0) ^∞ (x^(2i+1) /(1−x^(2i+1) )).........(1) from Ω=∫_0 ^1 ψ(x)+ψ((x)^(1/3) )+ψ((x)^(1/5) )+ψ((x)^(1/7) )((ln((1/x)))/x)dx Ω=−∫_0 ^1 [ψ(x)+ψ(x^(1/3) )+ψ(x^(1/5) )+ψ(x^(1/7) )]((lnx)/x)dx there the series form be Ω=∫_0 ^1 [Σ_(n=0) ^∞ ψ(x^(1/(2n+1)) )((lnx)/x)]dx.........(2) putting equation (1) into (2) Ω=−∫_0 ^1 Σ_(n=0) ^∞ Σ_(i=0) ^∞ [(((x^(2i+1) .x^(1/(2n+1)) )/(1−x^(2i+1) .x^(1/(2n+1)) )))((lnx)/x)]dx Ω=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ ∫_0 ^1 [((x^((2i+1)/(2n+1)) /(1−x^((2i+1)/(2n+1)) )))((lnx)/x)]dx let y=((2i+1)/(2n+1)) Ω=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ ∫_0 ^1 [((x^y /(1−x^y )))((lnx)/x)]dx=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ ∫_0 ^1 ((x^(y−1) /(1−x^y )))lnxdx let the series form of (1/(1−x^y ))=Σ_(m=0) ^∞ x^(my) Ω=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ ∫_0 ^1 (Σ_(m=0) ^∞ x^(my) .x^(y−1) lnx)dx (∂/∂a)∣_(a=0) Ω(a)=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ Σ_(m=0) ^∞ ∫_0 ^1 (x^(my) .x^(y−1) .x^a )dx (∂/∂a)∣_(a=0) Ω(a)=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ Σ_(m=0) ^∞ ∫_0 ^1 x^(my+y+a−1) dx (∂/∂a)∣_(a=0) Ω(a)=−Σ_(n=0) ^∞ Σ_(i=0) ^∞ Σ_(m=0) ^∞ ((1/(my+y+a))) Ω^′ (0)=Σ_(n=0) ^∞ Σ_(i=0) ^∞ Σ_(m=0) ^∞ [(1/((my+y)^2 ))]=(Σ_(n=0) ^∞ Σ_(i=0) ^∞ )(1/y^2 )Σ_(m=0) ^∞ (1/((1+m)^2 )) let z=1+m at m=0 ,z=1 Ω(0)=(Σ_(n=0) ^∞ Σ_(i=0) ^∞ )(1/y^2 )Σ_(z=1) ^∞ (1/z^2 )=(Σ_(n=0) ^∞ Σ_(i=0) ^∞ )(1/y^2 )ζ(2)=(π^2 /6)(Σ_(n=0) ^∞ Σ_(i=0) ^∞ )(1/y^2 ) but y=((2i+1)/(2n+1)) Ω(0)=(π^2 /6)(Σ_(n=0) ^∞ Σ_(i=0) ^∞ (1/((((2i+1)/(2n+1)))^2 )))=(π^2 /6)Σ_(n=0) ^∞ (2n+1)^2 Σ_(i=0) ^∞ (1/((2i+1)^2 )) Ω=(π^2 /6)[1^2 +3^2 +5^2 +7^2 ]×[(1/1^2 )+(1/3^2 )+(1/5^2 )+(1/7^2 )] Ω=(π^2 /6)(1+9+25+49)×[(1/1)+(1/9)+(1/(25))+(1/(49))] Ω=(π^2 /6)•(84)•((12916)/(11025))=((25832π^2 )/(1575)) ∵∫_0 ^1 [ψ(x)+ψ((x)^(1/3) )+ψ((x)^(1/5) )+ψ((x)^(1/7) )]((ln((1/x)))/x)=((25832π^2 )/(1575)) by mathew monday(05/09/2020)

$${if}\: \\ $$$$\psi\left({x}\right)=\frac{{x}}{\mathrm{1}−{x}}+\frac{{x}^{\mathrm{3}} }{\mathrm{1}−{x}^{\mathrm{2}} }+\frac{{x}^{\mathrm{5}} }{\mathrm{1}−{x}^{\mathrm{5}} }+\frac{{x}^{\mathrm{7}} }{\mathrm{1}−{x}^{\mathrm{7}} } \\ $$$${show}\:{that}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left[\psi\left({x}\right)+\psi\left(\sqrt[{\mathrm{3}}]{{x}}\right)+\psi\left(\sqrt[{\mathrm{5}}]{{x}}\right)+\psi\left(\sqrt[{\mathrm{7}}]{{x}}\right)\right]\frac{\mathrm{ln}\left(\frac{\mathrm{1}}{{x}}\right)}{{x}}=\frac{\mathrm{25832}\pi^{\mathrm{2}} }{\mathrm{1575}} \\ $$$${soltion}\: \\ $$$${let}\:{the}\:{generating}\:{series}\:{form}\:{of} \\ $$$$\psi\left({x}\right)=\frac{{x}}{\mathrm{1}−{x}}+\frac{{x}^{\mathrm{3}} }{\mathrm{1}−{x}^{\mathrm{2}} }+\frac{{x}^{\mathrm{5}} }{\mathrm{1}−{x}^{\mathrm{5}} }+\frac{{x}^{\mathrm{7}} }{\mathrm{1}−{x}^{\mathrm{7}} } \\ $$$${be}\:\psi\left({x}\right)=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{i}+\mathrm{1}} }{\mathrm{1}−{x}^{\mathrm{2}{i}+\mathrm{1}} }.........\left(\mathrm{1}\right) \\ $$$${from}\:\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} \psi\left({x}\right)+\psi\left(\sqrt[{\mathrm{3}}]{{x}}\right)+\psi\left(\sqrt[{\mathrm{5}}]{{x}}\right)+\psi\left(\sqrt[{\mathrm{7}}]{{x}}\right)\frac{\mathrm{ln}\left(\frac{\mathrm{1}}{{x}}\right)}{{x}}{dx} \\ $$$$\Omega=−\int_{\mathrm{0}} ^{\mathrm{1}} \left[\psi\left({x}\right)+\psi\left({x}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)+\psi\left({x}^{\frac{\mathrm{1}}{\mathrm{5}}} \right)+\psi\left({x}^{\frac{\mathrm{1}}{\mathrm{7}}} \right)\right]\frac{\mathrm{ln}{x}}{{x}}{dx} \\ $$$${there}\:{the}\:{series}\:{form}\:{be} \\ $$$$\Omega=\int_{\mathrm{0}} ^{\mathrm{1}} \left[\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\psi\left({x}^{\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}} \right)\frac{\mathrm{ln}{x}}{{x}}\right]{dx}.........\left(\mathrm{2}\right) \\ $$$${putting}\:{equation}\:\left(\mathrm{1}\right)\:{into}\:\left(\mathrm{2}\right) \\ $$$$\Omega=−\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\left(\frac{{x}^{\mathrm{2}{i}+\mathrm{1}} .{x}^{\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}} }{\mathrm{1}−{x}^{\mathrm{2}{i}+\mathrm{1}} .{x}^{\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}} }\right)\frac{\mathrm{ln}{x}}{{x}}\right]{dx} \\ $$$$\Omega=−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \left[\left(\frac{{x}^{\frac{\mathrm{2}{i}+\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}} }{\mathrm{1}−{x}^{\frac{\mathrm{2}{i}+\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}} }\right)\frac{\mathrm{ln}{x}}{{x}}\right]{dx} \\ $$$${let}\:{y}=\frac{\mathrm{2}{i}+\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\Omega=−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \left[\left(\frac{{x}^{{y}} }{\mathrm{1}−{x}^{{y}} }\right)\frac{\mathrm{ln}{x}}{{x}}\right]{dx}=−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{{x}^{{y}−\mathrm{1}} }{\mathrm{1}−{x}^{{y}} }\right)\mathrm{ln}{xdx} \\ $$$${let}\:{the}\:{series}\:{form}\:{of}\:\:\:\frac{\mathrm{1}}{\mathrm{1}−{x}^{{y}} }=\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{my}} \\ $$$$\Omega=−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{my}} .{x}^{{y}−\mathrm{1}} \mathrm{ln}{x}\right){dx} \\ $$$$\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{0}} \Omega\left({a}\right)=−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}^{{my}} .{x}^{{y}−\mathrm{1}} .{x}^{{a}} \right){dx} \\ $$$$\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{0}} \Omega\left({a}\right)=−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{my}+{y}+{a}−\mathrm{1}} {dx} \\ $$$$\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{0}} \Omega\left({a}\right)=−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{my}+{y}+{a}}\right) \\ $$$$\Omega^{'} \left(\mathrm{0}\right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\frac{\mathrm{1}}{\left({my}+{y}\right)^{\mathrm{2}} }\right]=\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\right)\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{1}+{m}\right)^{\mathrm{2}} } \\ $$$${let}\:{z}=\mathrm{1}+{m}\:{at}\:{m}=\mathrm{0}\:\:,{z}=\mathrm{1} \\ $$$$\Omega\left(\mathrm{0}\right)=\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\right)\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\underset{{z}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{z}^{\mathrm{2}} }=\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\right)\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\zeta\left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\right)\frac{\mathrm{1}}{{y}^{\mathrm{2}} } \\ $$$${but}\:{y}=\frac{\mathrm{2}{i}+\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\Omega\left(\mathrm{0}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\frac{\mathrm{2}{i}+\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\right)^{\mathrm{2}} }\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} \underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{i}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Omega=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\left[\mathrm{1}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} \right]×\left[\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{2}} }\right] \\ $$$$\Omega=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\left(\mathrm{1}+\mathrm{9}+\mathrm{25}+\mathrm{49}\right)×\left[\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{25}}+\frac{\mathrm{1}}{\mathrm{49}}\right] \\ $$$$\Omega=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\bullet\left(\mathrm{84}\right)\bullet\frac{\mathrm{12916}}{\mathrm{11025}}=\frac{\mathrm{25832}\pi^{\mathrm{2}} }{\mathrm{1575}} \\ $$$$\because\int_{\mathrm{0}} ^{\mathrm{1}} \left[\psi\left({x}\right)+\psi\left(\sqrt[{\mathrm{3}}]{{x}}\right)+\psi\left(\sqrt[{\mathrm{5}}]{{x}}\right)+\psi\left(\sqrt[{\mathrm{7}}]{{x}}\right)\right]\frac{\mathrm{ln}\left(\frac{\mathrm{1}}{{x}}\right)}{{x}}=\frac{\mathrm{25832}\pi^{\mathrm{2}} }{\mathrm{1575}} \\ $$$${by}\:{mathew}\:{monday}\left(\mathrm{05}/\mathrm{09}/\mathrm{2020}\right) \\ $$$$ \\ $$

Question Number 111799    Answers: 3   Comments: 4

Question Number 111788    Answers: 1   Comments: 0

Question Number 111781    Answers: 0   Comments: 3

Question Number 111774    Answers: 1   Comments: 0

Question Number 111772    Answers: 1   Comments: 0

calculate lim_(n→+∞) Π_(k=1) ^n (1+((√(k(n−k)))/n^2 ))

$$\mathrm{calculate}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\prod_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\left(\mathrm{1}+\frac{\sqrt{\mathrm{k}\left(\mathrm{n}−\mathrm{k}\right)}}{\mathrm{n}^{\mathrm{2}} }\right) \\ $$

Question Number 111771    Answers: 0   Comments: 0

calculate lim_(n→+∞) a_n ∫_0 ^1 x^(2n) sin(((πx)/2))dx with a_n =Σ_(k=1) ^n sin(((πk)/(2n)))

$$\mathrm{calculate}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{a}_{\mathrm{n}} \int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{x}^{\mathrm{2n}} \mathrm{sin}\left(\frac{\pi\mathrm{x}}{\mathrm{2}}\right)\mathrm{dx}\:\mathrm{with}\:\mathrm{a}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{sin}\left(\frac{\pi\mathrm{k}}{\mathrm{2n}}\right) \\ $$

Question Number 111770    Answers: 0   Comments: 0

f function continue on [0,1] find lim_(n→+∞) (1/n)Σ_(k=0) ^n (n−k)∫_(k/n) ^((k+1)/n) f(x)dx

$$\mathrm{f}\:\mathrm{function}\:\mathrm{continue}\:\mathrm{on}\:\left[\mathrm{0},\mathrm{1}\right]\:\mathrm{find}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\frac{\mathrm{1}}{\mathrm{n}}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \left(\mathrm{n}−\mathrm{k}\right)\int_{\frac{\mathrm{k}}{\mathrm{n}}} ^{\frac{\mathrm{k}+\mathrm{1}}{\mathrm{n}}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ $$

Question Number 111769    Answers: 0   Comments: 0

let u_n =Σ_(k=1) ^n (1/(√((k+n)(k+n+1)))) detremine lim_(n→+∞) u_n

$$\mathrm{let}\:\mathrm{u}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\sqrt{\left(\mathrm{k}+\mathrm{n}\right)\left(\mathrm{k}+\mathrm{n}+\mathrm{1}\right)}} \\ $$$$\mathrm{detremine}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{u}_{\mathrm{n}} \\ $$

Question Number 111768    Answers: 0   Comments: 0

explicite f(x) =∫_0 ^(2π) ln(x^2 −2xcosθ +1)dθ (x≠+^− 1)

$$\mathrm{explicite}\:\mathrm{f}\left(\mathrm{x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2xcos}\theta\:+\mathrm{1}\right)\mathrm{d}\theta\:\:\:\:\:\:\:\:\left(\mathrm{x}\neq\overset{−} {+}\mathrm{1}\right) \\ $$

Question Number 111766    Answers: 2   Comments: 0

calculate lim_(n→+∞) Σ_(k=1) ^n ln((n/(n+k)))^(1/n)

$$\mathrm{calculate}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \mathrm{ln}\left(\frac{\mathrm{n}}{\mathrm{n}+\mathrm{k}}\right)^{\frac{\mathrm{1}}{\mathrm{n}}} \\ $$

Question Number 111762    Answers: 1   Comments: 0

caoculate ∫_0 ^(π/4) ln(1+2tanx)dx

$$\mathrm{caoculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{1}+\mathrm{2tanx}\right)\mathrm{dx} \\ $$

Question Number 111760    Answers: 0   Comments: 0

find lim_(x→1^+ ) ∫_x ^x^2 ((ln(t))/((t−1)^2 ))dx

$$\mathrm{find}\:\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{1}^{+} } \:\:\:\int_{\mathrm{x}} ^{\mathrm{x}^{\mathrm{2}} } \:\:\frac{\mathrm{ln}\left(\mathrm{t}\right)}{\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx} \\ $$

Question Number 111756    Answers: 0   Comments: 0

find I_n =∫_0 ^(π/4) (du/(cos^n u))

$$\mathrm{find}\:\mathrm{I}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{\mathrm{du}}{\mathrm{cos}^{\mathrm{n}} \mathrm{u}} \\ $$

Question Number 111755    Answers: 2   Comments: 0

calculate lim_(n→+∞) Σ_(k=1) ^n (√((n−k)/(n^3 +n^2 k)))

$$\mathrm{calculate}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\:\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\:\sqrt{\frac{\mathrm{n}−\mathrm{k}}{\mathrm{n}^{\mathrm{3}} \:+\mathrm{n}^{\mathrm{2}} \mathrm{k}}} \\ $$

Question Number 111754    Answers: 2   Comments: 0

calculate lim_(n→+∞) Σ_(k=1) ^(2n) (k/(k^2 +n^2 ))

$$ \\ $$$$\mathrm{calculate}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{2n}} \:\:\:\:\frac{\mathrm{k}}{\mathrm{k}^{\mathrm{2}} \:+\mathrm{n}^{\mathrm{2}} } \\ $$

Question Number 111887    Answers: 0   Comments: 0

Question Number 111882    Answers: 1   Comments: 0

(√(bemath)) (1)lim_(x→∞) [(x^2 /(x+1)) − (x^2 /(x+3)) ] ? (2) prove that n^2 ≤ 2^n for ∀n∈N by mathematical induction

$$\:\:\sqrt{{bemath}} \\ $$$$\left(\mathrm{1}\right)\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[\frac{{x}^{\mathrm{2}} }{{x}+\mathrm{1}}\:−\:\frac{{x}^{\mathrm{2}} }{{x}+\mathrm{3}}\:\right]\:? \\ $$$$\left(\mathrm{2}\right)\:{prove}\:{that}\:{n}^{\mathrm{2}} \:\leqslant\:\mathrm{2}^{{n}} \:{for}\:\forall{n}\in\mathbb{N} \\ $$$${by}\:{mathematical}\:{induction} \\ $$

Question Number 111745    Answers: 0   Comments: 6

Question Number 111826    Answers: 0   Comments: 2

There are two cards; one is yellow on both sides and the other one is yellow in one side and red on the other. The cards have the same probability ((1/2)) of being chosen, and one is chosen and placed on the table. If the upper side of the card on the table is yellow, then the probability that the under side is also yellow is.

$$\mathrm{There}\:\mathrm{are}\:\mathrm{two}\:\mathrm{cards};\:\mathrm{one}\:\mathrm{is}\:\mathrm{yellow}\:\mathrm{on} \\ $$$$\mathrm{both}\:\mathrm{sides}\:\mathrm{and}\:\mathrm{the}\:\mathrm{other}\:\mathrm{one}\:\mathrm{is}\:\mathrm{yellow} \\ $$$$\mathrm{in}\:\mathrm{one}\:\mathrm{side}\:\mathrm{and}\:\mathrm{red}\:\mathrm{on}\:\mathrm{the}\:\mathrm{other}.\:\mathrm{The} \\ $$$$\mathrm{cards}\:\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{probability}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\mathrm{of}\:\mathrm{being}\:\mathrm{chosen},\:\mathrm{and}\:\mathrm{one}\:\mathrm{is}\:\mathrm{chosen}\:\mathrm{and} \\ $$$$\mathrm{placed}\:\mathrm{on}\:\mathrm{the}\:\mathrm{table}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{upper}\:\mathrm{side} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{card}\:\mathrm{on}\:\mathrm{the}\:\mathrm{table}\:\mathrm{is}\:\mathrm{yellow}, \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{the}\:\mathrm{under} \\ $$$$\mathrm{side}\:\mathrm{is}\:\mathrm{also}\:\mathrm{yellow}\:\mathrm{is}. \\ $$

Question Number 111734    Answers: 1   Comments: 0

A chord which is a perpendicular bisector of radius of length 18cm in a circle, has length.

$$\mathrm{A}\:\mathrm{chord}\:\mathrm{which}\:\mathrm{is}\:\mathrm{a}\:\mathrm{perpendicular} \\ $$$$\mathrm{bisector}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{length}\:\mathrm{18cm}\:\mathrm{in}\:\mathrm{a} \\ $$$$\mathrm{circle},\:\mathrm{has}\:\mathrm{length}. \\ $$$$ \\ $$

Question Number 111732    Answers: 1   Comments: 0

A blind man is to place 5 letters into 5 pigeon holes, how many ways can 4 of the letters be wrongly placed? (note that only one letter must be in a pigeon hole)

$$\mathrm{A}\:\mathrm{blind}\:\mathrm{man}\:\mathrm{is}\:\mathrm{to}\:\mathrm{place}\:\mathrm{5}\:\mathrm{letters}\:\mathrm{into}\:\mathrm{5} \\ $$$$\mathrm{pigeon}\:\mathrm{holes},\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{4}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{letters}\:\mathrm{be}\:\mathrm{wrongly}\:\mathrm{placed}? \\ $$$$\left(\mathrm{note}\:\mathrm{that}\:\mathrm{only}\:\mathrm{one}\:\mathrm{letter}\:\mathrm{must}\:\mathrm{be}\:\mathrm{in}\:\mathrm{a}\right. \\ $$$$\left.\mathrm{pigeon}\:\mathrm{hole}\right) \\ $$

Question Number 111730    Answers: 0   Comments: 9

How many triples of positive integers (x,y,z) satisfy 79x+80y+81z =2016

$$\mathrm{How}\:\mathrm{many}\:\mathrm{triples}\:\mathrm{of}\:\mathrm{positive}\:\mathrm{integers} \\ $$$$\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\:\mathrm{satisfy}\: \\ $$$$\mathrm{79x}+\mathrm{80y}+\mathrm{81z}\:=\mathrm{2016} \\ $$

Question Number 112813    Answers: 0   Comments: 2

A binary operation has the property a∗(b∗c) = (a∗b)•c and that a∗a=1 for all non−zero real numbers a,b and c. (′•′ here represent multiplication). The solution of the equation 2016∗(6∗x)=100 can be written as (p/q) where p and q are relatively prime positive integers. What is q−p?

$$\mathrm{A}\:\mathrm{binary}\:\mathrm{operation}\:\mathrm{has}\:\mathrm{the}\:\mathrm{property} \\ $$$$\mathrm{a}\ast\left(\mathrm{b}\ast\mathrm{c}\right)\:=\:\left(\mathrm{a}\ast\mathrm{b}\right)\bullet\mathrm{c}\:\mathrm{and}\:\mathrm{that}\:\mathrm{a}\ast\mathrm{a}=\mathrm{1}\:\mathrm{for} \\ $$$$\mathrm{all}\:\mathrm{non}−\mathrm{zero}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{a},\mathrm{b}\:\mathrm{and}\:\mathrm{c}. \\ $$$$\left('\bullet'\:\mathrm{here}\:\mathrm{represent}\:\mathrm{multiplication}\right). \\ $$$$\mathrm{The}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{2016}\ast\left(\mathrm{6}\ast\mathrm{x}\right)=\mathrm{100}\:\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{as}\:\frac{\mathrm{p}}{\mathrm{q}} \\ $$$$\mathrm{where}\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}\:\mathrm{are}\:\mathrm{relatively}\:\mathrm{prime} \\ $$$$\mathrm{positive}\:\mathrm{integers}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{q}−\mathrm{p}? \\ $$

Question Number 112812    Answers: 0   Comments: 2

There are 2016 straight lines drawn on a board such that (1/2) of the lines are parallel to one another. (3/8) of them meet at a point and each of the remaining ones intersect with all other lines on the board. Determine the total number of intersections possible.

$$\mathrm{There}\:\mathrm{are}\:\mathrm{2016}\:\mathrm{straight}\:\mathrm{lines}\:\mathrm{drawn}\:\mathrm{on} \\ $$$$\mathrm{a}\:\mathrm{board}\:\mathrm{such}\:\mathrm{that}\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{of}\:\mathrm{the}\:\mathrm{lines}\:\mathrm{are} \\ $$$$\mathrm{parallel}\:\mathrm{to}\:\mathrm{one}\:\mathrm{another}.\:\frac{\mathrm{3}}{\mathrm{8}}\:\mathrm{of}\:\mathrm{them} \\ $$$$\mathrm{meet}\:\mathrm{at}\:\mathrm{a}\:\mathrm{point}\:\mathrm{and}\:\mathrm{each}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{remaining}\:\mathrm{ones}\:\mathrm{intersect}\:\mathrm{with}\:\mathrm{all} \\ $$$$\mathrm{other}\:\mathrm{lines}\:\mathrm{on}\:\mathrm{the}\:\mathrm{board}.\:\mathrm{Determine} \\ $$$$\mathrm{the}\:\mathrm{total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{intersections} \\ $$$$\mathrm{possible}. \\ $$

Question Number 111725    Answers: 1   Comments: 0

Find the positive integer n such that tan^(−1) ((1/3))+tan^(−1) ((1/4))+tan^(−1) ((1/5))+tan^(−1) ((1/n))=(π/4)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{n}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{5}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{n}}\right)=\frac{\pi}{\mathrm{4}} \\ $$

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