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Question Number 111923    Answers: 2   Comments: 0

(√(bemath )) lim_(x→0) ((((cos 4x))^(1/(3 )) −1)/(cos 3x−cos 9x)) ?

$$\sqrt{{bemath}\:} \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{cos}\:\mathrm{4}{x}}\:−\mathrm{1}}{\mathrm{cos}\:\mathrm{3}{x}−\mathrm{cos}\:\mathrm{9}{x}}\:? \\ $$

Question Number 111982    Answers: 2   Comments: 0

Question Number 111914    Answers: 1   Comments: 0

lim_(x→0) ((8x^3 )/( (√(4+sin 6x)) −(√(tan 6x+4)))) ?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{8}{x}^{\mathrm{3}} }{\:\sqrt{\mathrm{4}+\mathrm{sin}\:\mathrm{6}{x}}\:−\sqrt{\mathrm{tan}\:\mathrm{6}{x}+\mathrm{4}}}\:? \\ $$

Question Number 111909    Answers: 2   Comments: 0

solve { ((x(√x)+y(√y) = 5)),((x(√y) +y(√x) = 1)) :}

$${solve}\:\begin{cases}{{x}\sqrt{{x}}+{y}\sqrt{{y}}\:=\:\mathrm{5}}\\{{x}\sqrt{{y}}\:+{y}\sqrt{{x}}\:=\:\mathrm{1}}\end{cases} \\ $$

Question Number 111907    Answers: 2   Comments: 0

lim_(x→(π/(20))) ((1−tan 5x)/(sin 5x−cos 5x)) ?

$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{20}}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{tan}\:\mathrm{5}{x}}{\mathrm{sin}\:\mathrm{5}{x}−\mathrm{cos}\:\mathrm{5}{x}}\:? \\ $$

Question Number 111906    Answers: 1   Comments: 0

Question Number 111890    Answers: 0   Comments: 0

Question Number 111879    Answers: 1   Comments: 0

prove that lim_(n→∞) nΣ_(k=1) ^(n−1) ((ln(k+n)−ln(n))/(k^2 +n^2 ))=(π/8)ln2

$${prove}\:{that}\: \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{n}\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{ln}\left({k}+{n}\right)−\mathrm{ln}\left({n}\right)}{{k}^{\mathrm{2}} +{n}^{\mathrm{2}} }=\frac{\pi}{\mathrm{8}}\mathrm{ln2} \\ $$

Question Number 111876    Answers: 0   Comments: 0

Question Number 111873    Answers: 1   Comments: 0

(√(bemath)) ∫ ((2−cos x)/(2+cos x)) dx

$$\:\:\:\sqrt{{bemath}} \\ $$$$\int\:\frac{\mathrm{2}−\mathrm{cos}\:{x}}{\mathrm{2}+\mathrm{cos}\:{x}}\:{dx}\: \\ $$

Question Number 111869    Answers: 0   Comments: 0

evaluate ∫_0 ^1 ((ln(1+x))/(x^2 +5x+6))dx

$${evaluate} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{x}\right)}{{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{6}}{dx} \\ $$

Question Number 111868    Answers: 2   Comments: 0

simplify ∫((1−5sin^2 x)/(cos^5 xsin^2 x))dx

$${simplify} \\ $$$$\int\frac{\mathrm{1}−\mathrm{5sin}^{\mathrm{2}} {x}}{\mathrm{cos}^{\mathrm{5}} {x}\mathrm{sin}^{\mathrm{2}} {x}}{dx} \\ $$

Question Number 111859    Answers: 2   Comments: 7

I=∫(dx/((x^2 +2x+3)(√(x^2 +x+3)))) = ? my try..

$${I}=\int\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\right)\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{3}}}\:=\:? \\ $$$${my}\:{try}.. \\ $$

Question Number 111850    Answers: 4   Comments: 0

If ∣z∣ = 3 , what is the maximum and minimum value of ∣z−1+i(√3) ∣ ?

$${If}\:\mid{z}\mid\:=\:\mathrm{3}\:,\:{what}\:{is}\:{the}\:{maximum} \\ $$$${and}\:{minimum}\:{value}\:{of}\:\mid{z}−\mathrm{1}+{i}\sqrt{\mathrm{3}}\:\mid\:? \\ $$

Question Number 112531    Answers: 0   Comments: 4

A rectangular cardboard is 8cm long and 6cm wide. What is the least number of beads you can arrange on the board such that there are at least two of the beads that are less than (√(10))cm apart.

$$\mathrm{A}\:\mathrm{rectangular}\:\mathrm{cardboard}\:\mathrm{is}\:\mathrm{8cm}\:\mathrm{long} \\ $$$$\mathrm{and}\:\mathrm{6cm}\:\mathrm{wide}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{least} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{beads}\:\mathrm{you}\:\mathrm{can}\:\mathrm{arrange}\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{board}\:\mathrm{such}\:\mathrm{that}\:\mathrm{there}\:\mathrm{are}\:\mathrm{at}\:\mathrm{least} \\ $$$$\mathrm{two}\:\mathrm{of}\:\mathrm{the}\:\mathrm{beads}\:\mathrm{that}\:\mathrm{are}\:\mathrm{less}\:\mathrm{than} \\ $$$$\sqrt{\mathrm{10}}\mathrm{cm}\:\mathrm{apart}. \\ $$

Question Number 111831    Answers: 1   Comments: 0

Let N be the greatest multiple of 36 all of whose digits are even and no two of whose digits are the same. Find the remainder when N is divided by 1000.

$$\mathrm{Let}\:\mathrm{N}\:\mathrm{be}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{36}\:\mathrm{all} \\ $$$$\mathrm{of}\:\mathrm{whose}\:\mathrm{digits}\:\mathrm{are}\:\mathrm{even}\:\mathrm{and}\:\mathrm{no}\:\mathrm{two}\:\mathrm{of} \\ $$$$\mathrm{whose}\:\mathrm{digits}\:\mathrm{are}\:\mathrm{the}\:\mathrm{same}.\:\mathrm{Find}\:\mathrm{the} \\ $$$$\mathrm{remainder}\:\mathrm{when}\:\mathrm{N}\:\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{1000}. \\ $$

Question Number 112535    Answers: 1   Comments: 5

Let K be the product of all factors (b−a) (not necessarily distinct) where a and b are integers satisfying 1≤a≤b≤10. Find the greatest integer n such that 2^n divides K.

$$\mathrm{Let}\:\mathrm{K}\:\mathrm{be}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{all}\:\mathrm{factors} \\ $$$$\left(\mathrm{b}−\mathrm{a}\right)\:\left(\mathrm{not}\:\mathrm{necessarily}\:\mathrm{distinct}\right) \\ $$$$\mathrm{where}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{are}\:\mathrm{integers}\:\mathrm{satisfying} \\ $$$$\mathrm{1}\leqslant\mathrm{a}\leqslant\mathrm{b}\leqslant\mathrm{10}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{greatest} \\ $$$$\mathrm{integer}\:\mathrm{n}\:\mathrm{such}\:\mathrm{that}\:\mathrm{2}^{\mathrm{n}} \:\mathrm{divides}\:\mathrm{K}. \\ $$$$ \\ $$

Question Number 111818    Answers: 3   Comments: 0

(√(bemath )) (1)∫ ((cos x)/(2−cos x)) dx (2) f(x) = ∣x^3 ∣ ⇒ f ′(x) ?

$$\:\:\:\sqrt{{bemath}\:} \\ $$$$\left(\mathrm{1}\right)\int\:\frac{\mathrm{cos}\:{x}}{\mathrm{2}−\mathrm{cos}\:{x}}\:{dx}\: \\ $$$$\left(\mathrm{2}\right)\:{f}\left({x}\right)\:=\:\mid{x}^{\mathrm{3}} \mid\:\Rightarrow\:{f}\:'\left({x}\right)\:? \\ $$

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

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

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