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Question Number 112533    Answers: 2   Comments: 2

Find the minimum number of n integers to be selected from S={1,2,3,...11} so that the difference of two of the n integers is 7.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{number}\:\mathrm{of}\:\mathrm{n} \\ $$$$\mathrm{integers}\:\mathrm{to}\:\mathrm{be}\:\mathrm{selected}\:\mathrm{from} \\ $$$$\mathrm{S}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},...\mathrm{11}\right\}\:\mathrm{so}\:\mathrm{that}\:\mathrm{the}\:\mathrm{difference} \\ $$$$\mathrm{of}\:\mathrm{two}\:\mathrm{of}\:\mathrm{the}\:\mathrm{n}\:\mathrm{integers}\:\mathrm{is}\:\mathrm{7}. \\ $$

Question Number 112060    Answers: 1   Comments: 0

In a trapezium, ABCD, with AB parallel to CD. If M is the midpoint of line segment AD and P is a point on line BC such that MP is perpendicular to BC. Show that, we need only the lengths of line segments MP and BC to calculate the area ABCD.

$$\mathrm{In}\:\mathrm{a}\:\mathrm{trapezium},\:\mathrm{ABCD},\:\mathrm{with}\:\mathrm{AB} \\ $$$$\mathrm{parallel}\:\mathrm{to}\:\mathrm{CD}.\:\mathrm{If}\:\mathrm{M}\:\mathrm{is}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of} \\ $$$$\mathrm{line}\:\mathrm{segment}\:\mathrm{AD}\:\mathrm{and}\:\mathrm{P}\:\mathrm{is}\:\mathrm{a}\:\mathrm{point}\:\mathrm{on} \\ $$$$\mathrm{line}\:\mathrm{BC}\:\mathrm{such}\:\mathrm{that}\:\mathrm{MP}\:\mathrm{is}\:\mathrm{perpendicular} \\ $$$$\mathrm{to}\:\mathrm{BC}.\:\mathrm{Show}\:\mathrm{that},\:\mathrm{we}\:\mathrm{need}\:\mathrm{only}\:\mathrm{the} \\ $$$$\mathrm{lengths}\:\mathrm{of}\:\mathrm{line}\:\mathrm{segments}\:\mathrm{MP}\:\mathrm{and}\:\mathrm{BC} \\ $$$$\mathrm{to}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{area}\:\mathrm{ABCD}. \\ $$

Question Number 111926    Answers: 1   Comments: 1

Question Number 112001    Answers: 0   Comments: 2

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

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