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answer to26109 S_n = Σ_(k=1) ^(k=n) (1/(k^2 (k+1)^2 )) we decompose F(X) = (1/(X^2 (X+1)^2 )) = (a/X) +(b/X^2^ ) +(c/(X+1)) +(d/((X+1)^2 )) we find F(X) = ((−2)/X) +(1/X^2 ) + (2/(X+1)) + (1/((X+1)^2 )) so S_n = −2Σ_(k=1) ^(k=n) (1/k) +2Σ_(k=1) ^(k=n) (1/(k+1)) +Σ_(k=1) ^(k=n) (1/k^2^ ) + Σ_(k=1) ^(k=n) (1/((k+1)^2 )) but Σ_(k=1) ^(k=n) (1/k) = H_n Σ_(k=1) ^(k=n) (1/(k+1))= H_(n+1) −1 Σ_(k=1) ^(k=n) (1/((k+1)^2 )) = Σ_(k=1) ^(k=n) (1/k^2 ) + (1/((n+1)^2 )) −1⇒ S_n = 2(H_(n+1) −H_n ) +2 Σ_(k=1) ^(k=n) (1/k^2 ) −3 but lim_(n−>∝) (H_(n+1) − H_n ) =0 and Σ_(k=1) ^∝ (1/k^2 ) = (π^2 /6) ⇒ lim_(n−>∝) S_n = 2 (π^2 /6) −3 = (π^2 /3) −3 .

$${answer}\:{to}\mathrm{26109}\:\:\:{S}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:\:{we}\:{decompose} \\ $$$${F}\left({X}\right)\:\:=\:\:\frac{\mathrm{1}}{{X}^{\mathrm{2}} \left({X}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:\:\frac{{a}}{{X}}\:\:\:+\frac{{b}}{{X}^{\mathrm{2}^{} } }\:\:+\frac{{c}}{{X}+\mathrm{1}}\:\:+\frac{{d}}{\left({X}+\mathrm{1}\right)^{\mathrm{2}} }\:\:{we}\:{find} \\ $$$${F}\left({X}\right)\:\:=\:\:\frac{−\mathrm{2}}{{X}}\:\:+\frac{\mathrm{1}}{{X}^{\mathrm{2}} }\:\:+\:\frac{\mathrm{2}}{{X}+\mathrm{1}}\:\:+\:\:\frac{\mathrm{1}}{\left({X}+\mathrm{1}\right)^{\mathrm{2}} }\:\:{so} \\ $$$$\:\:{S}_{{n}} \:\:=\:\:−\mathrm{2}\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\:\frac{\mathrm{1}}{{k}}\:\:+\mathrm{2}\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:\:+\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}^{} } }\:\:+\:\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${but}\:\:\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\frac{\mathrm{1}}{{k}}\:\:=\:{H}_{{n}} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\:\frac{\mathrm{1}}{{k}+\mathrm{1}}=\:\:{H}_{{n}+\mathrm{1}} \:\:−\mathrm{1} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} }\:\:=\:\:\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:\:+\:\:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\:\:−\mathrm{1}\Rightarrow \\ $$$${S}_{{n}} \:\:=\:\:\mathrm{2}\left({H}_{{n}+\mathrm{1}} \:\:−{H}_{{n}} \right)\:\:+\mathrm{2}\:\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:\:\:−\mathrm{3} \\ $$$${but}\:\:\:{lim}_{{n}−>\propto} \left({H}_{{n}+\mathrm{1}} −\:{H}_{{n}} \right)\:\:=\mathrm{0}\:\:\:{and}\:\:\:\:\sum_{{k}=\mathrm{1}} ^{\propto} \:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\Rightarrow\:\:\:{lim}_{{n}−>\propto} \:\:{S}_{{n}} \:\:=\:\mathrm{2}\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\:−\mathrm{3}\:\:=\:\:\frac{\pi^{\mathrm{2}} }{\mathrm{3}}\:\:−\mathrm{3}\:\:\:\:\:. \\ $$

Question Number 26117 Answers: 1 Comments: 0

(1/6)(√((3log1728)/(1+(1/2)log36+(1/3)log8))) simplify the question above

$$\frac{\mathrm{1}}{\mathrm{6}}\sqrt{\frac{\mathrm{3}{log}\mathrm{1728}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{log}\mathrm{36}+\frac{\mathrm{1}}{\mathrm{3}}{log}\mathrm{8}}} \\ $$$${simplify}\:{the}\:{question}\:{above} \\ $$

Question Number 26115 Answers: 1 Comments: 0

x^2 +(1/x^2 )=3 thrn find the valu of( x−(1/x))^2

$$\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=\mathrm{3}\:\mathrm{thrn}\:\mathrm{find}\:\mathrm{the}\:\mathrm{valu}\:\mathrm{of}\left(\:\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{2}} \\ $$

Question Number 26126 Answers: 0 Comments: 1

using 1st principle find the derivative of y=x^x

$${using}\:\mathrm{1}{st}\:{principle}\:{find}\:{the} \\ $$$${derivative}\:{of} \\ $$$$\:\:\:\:\:\:\:\:{y}={x}^{{x}} \\ $$

Question Number 26113 Answers: 1 Comments: 0

Question Number 26107 Answers: 0 Comments: 0

answer to 26024 let put c= ∫_0 ^∞ cos(ax^2 )dx and c = ∫_0 ^∞ sin(ax^2 )dx ew have c−is = ∫_0 ^∞ e^(−iax^2 ) dx =2^(−1) ∫_R e^(−iax^2 ) dx and i put x^(1/2) =r(x)(notation) so 2(c−is) = ∫_R e^(−(r(ia)x)^2 ) dx and by the changement t= r(ia) x we find 2(c+is) = (r(ia))^(−1) ∫_R e^(−t^2 ) dt = r(π)/r(ia) but r(ia) =r(i) r(a) = r(a) e^ −−>2(c+is) = r(π) r(a)^(−1) e^(−iπ/4) ^) −−> c = r(2π)/_(4r(a)) and s = r(2π)/_(4r(a))

$${answer}\:{to}\:\mathrm{26024}\:\:\:\:{let}\:\:{put}\:{c}=\:\int_{\mathrm{0}} ^{\infty} \:{cos}\left({ax}^{\mathrm{2}} \right){dx}\:\:\:{and}\:\:\:{c}\:=\:\:\int_{\mathrm{0}} ^{\infty} \:{sin}\left({ax}^{\mathrm{2}} \right){dx} \\ $$$${ew}\:{have}\:\:{c}−{is}\:\:=\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{iax}^{\mathrm{2}} } {dx}\:\:\:=\mathrm{2}^{−\mathrm{1}} \:\:\int_{{R}} \:{e}^{−{iax}^{\mathrm{2}} } {dx}\:\:\:{and}\:\:{i}\:{put} \\ $$$${x}^{\mathrm{1}/\mathrm{2}} \:={r}\left({x}\right)\left({notation}\right)\:\:{so}\:\:\mathrm{2}\left({c}−{is}\right)\:\:=\:\:\:\int_{{R}} \:\:{e}^{−\left({r}\left({ia}\right){x}\right)^{\mathrm{2}} } {dx} \\ $$$${and}\:{by}\:{the}\:{changement}\:\:\:{t}=\:{r}\left({ia}\right)\:{x}\:\:\:{we}\:{find} \\ $$$$\mathrm{2}\left({c}+{is}\right)\:\:=\:\:\:\left({r}\left({ia}\right)\right)^{−\mathrm{1}} \:\int_{{R}} \:{e}^{−{t}^{\mathrm{2}} } {dt}\:\:\:=\:\:{r}\left(\pi\right)/{r}\left({ia}\right)\:\:{but} \\ $$$${r}\left({ia}\right)\:\:={r}\left({i}\right)\:{r}\left({a}\right)\:\:=\:\:{r}\left({a}\right)\:{e}^{} \:\:\:−−>\mathrm{2}\left({c}+{is}\right)\:\:\:\:=\:\:{r}\left(\pi\right)\:{r}\left({a}\right)^{−\mathrm{1}} \:\:{e}^{−{i}\pi/\mathrm{4}} \:\:^{\left.\right)} \\ $$$$−−>\:\:{c}\:=\:\:\:{r}\left(\mathrm{2}\pi\right)/_{\mathrm{4}{r}\left({a}\right)} \:\:{and}\:\:\:{s}\:\:=\:\:{r}\left(\mathrm{2}\pi\right)/_{\mathrm{4}{r}\left({a}\right)} \\ $$

Question Number 26111 Answers: 0 Comments: 2

find the radius of convergence for the serie Σ_(n=1) ^∝ H_n x^n H_n = Σ_(k=1) ^(k=n) (1/k) .

$${find}\:\:{the}\:{radius}\:{of}\:{convergence}\:{for}\:{the}\:{serie}\:\:\:\sum_{{n}=\mathrm{1}} ^{\propto} \:{H}_{{n}} \:{x}^{{n}} \\ $$$${H}_{{n}} \:\:=\:\:\:\sum_{{k}=\mathrm{1}} ^{{k}={n}} \:\:\frac{\mathrm{1}}{{k}}\:. \\ $$

Question Number 26223 Answers: 1 Comments: 1

let put ξ(x)= Σ_(n=1) ^∝ (1/n^x ) with x>1 and δ(x) =Σ_(n=1) ^∝ (((−1)^n )/n^x ) find a relation between ξ(x) and δ(x).

$${let}\:{put}\:\xi\left({x}\right)=\:\:\sum_{{n}=\mathrm{1}} ^{\propto} \:\:\frac{\mathrm{1}}{{n}^{{x}} }\:\:{with}\:{x}>\mathrm{1} \\ $$$${and}\:\:\delta\left({x}\right)\:\:=\sum_{{n}=\mathrm{1}} ^{\propto} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{{x}} }\:\:\:{find}\:{a}\:{relation} \\ $$$${between}\:\xi\left({x}\right)\:{and}\:\delta\left({x}\right). \\ $$

Question Number 26098 Answers: 0 Comments: 0

Question Number 26090 Answers: 1 Comments: 0

∫((asin^3 θ+bcos^3 θ)/(sin^2 θ.cos^2 θ))dθ

$$\int\frac{{a}\mathrm{sin}\:^{\mathrm{3}} \theta+{b}\mathrm{cos}\:^{\mathrm{3}} \theta}{\mathrm{sin}^{\mathrm{2}} \:\theta.\mathrm{cos}^{\mathrm{2}} \:\theta}{d}\theta \\ $$

Question Number 26087 Answers: 0 Comments: 2

Given f(x) = (1 − x + x^2 − x^3 + ... − x^(2015) + x^(2016) )^2 Find the sum of all odd coeffisiens! Ex. f(x) = (x^2 + x + 1)^2 = 1x^4 + 2x^3 + 3x^2 + 2x + 1 The sum of odd coeffisien is 1 + 3 = 4

$$\mathrm{Given} \\ $$$${f}\left({x}\right)\:=\:\left(\mathrm{1}\:−\:{x}\:+\:{x}^{\mathrm{2}} \:−\:{x}^{\mathrm{3}} \:+\:...\:−\:{x}^{\mathrm{2015}} \:+\:{x}^{\mathrm{2016}} \right)^{\mathrm{2}} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{odd}\:\mathrm{coeffisiens}! \\ $$$$ \\ $$$$\mathrm{Ex}.\:{f}\left({x}\right)\:=\:\left({x}^{\mathrm{2}} \:+\:{x}\:+\:\mathrm{1}\right)^{\mathrm{2}} \:=\:\mathrm{1}{x}^{\mathrm{4}} \:+\:\mathrm{2}{x}^{\mathrm{3}} \:+\:\mathrm{3}{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:+\:\mathrm{1} \\ $$$$\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{odd}\:\mathrm{coeffisien}\:\mathrm{is}\:\mathrm{1}\:+\:\mathrm{3}\:=\:\mathrm{4} \\ $$

Question Number 26078 Answers: 1 Comments: 0

x^2 −x−42 factorise

$$\mathrm{x}^{\mathrm{2}} −\mathrm{x}−\mathrm{42}\:\mathrm{factorise} \\ $$

Question Number 26073 Answers: 0 Comments: 2

solve the differential equation(D^2 +2D+1)y=x^2 +2x+1

$$ \\ $$$${solve}\:{the}\:{differential}\:{equation}\left({D}^{\mathrm{2}} +\mathrm{2}{D}+\mathrm{1}\right){y}={x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1} \\ $$

Question Number 26067 Answers: 2 Comments: 0

Find the value of ((2 + 3^2 )/(1! + 2! + 3! + 4!)) + ((3 + 4^2 )/(2! + 3! + 4! + 5!)) + ... + ((2013 + 2014^2 )/(2012! + 2013! + 2014! + 2015!))

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\frac{\mathrm{2}\:+\:\mathrm{3}^{\mathrm{2}} }{\mathrm{1}!\:+\:\mathrm{2}!\:+\:\mathrm{3}!\:+\:\mathrm{4}!}\:+\:\frac{\mathrm{3}\:+\:\mathrm{4}^{\mathrm{2}} }{\mathrm{2}!\:+\:\mathrm{3}!\:+\:\mathrm{4}!\:+\:\mathrm{5}!}\:+\:...\:+\:\frac{\mathrm{2013}\:+\:\mathrm{2014}^{\mathrm{2}} }{\mathrm{2012}!\:+\:\mathrm{2013}!\:+\:\mathrm{2014}!\:+\:\mathrm{2015}!} \\ $$

Question Number 26109 Answers: 0 Comments: 1

let s give S_n = Σ_(k=1) ^(k=n) k^(−2 ) . (k+1)^(−2) find lim_(n−>∝) S_n .