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Question Number 165471 Answers: 0 Comments: 1

{ ((h(3x)=(((2−x)/(x+1))−f(x^3 ))^2 )),((f(1)=f ′(1)=2)) :} h ′(3)=?

$$\:\begin{cases}{{h}\left(\mathrm{3}{x}\right)=\left(\frac{\mathrm{2}−{x}}{{x}+\mathrm{1}}−{f}\left({x}^{\mathrm{3}} \right)\right)^{\mathrm{2}} }\\{{f}\left(\mathrm{1}\right)={f}\:'\left(\mathrm{1}\right)=\mathrm{2}}\end{cases} \\ $$$$\:{h}\:'\left(\mathrm{3}\right)=? \\ $$

Question Number 165457 Answers: 1 Comments: 19

Question Number 165514 Answers: 1 Comments: 0

Question Number 165512 Answers: 0 Comments: 0

Σ_(n=0) ^∞ ((2n+1)/(e^((2n+1)π) +1))=^? (1/(24)) −−−−−−−−−−−−−−by Mr. Levent

$$\underset{\boldsymbol{\mathrm{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}\boldsymbol{\mathrm{n}}+\mathrm{1}}{\boldsymbol{\mathrm{e}}^{\left(\mathrm{2}\boldsymbol{\mathrm{n}}+\mathrm{1}\right)\pi} +\mathrm{1}}\overset{?} {=}\frac{\mathrm{1}}{\mathrm{24}} \\ $$$$−−−−−−−−−−−−−−\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{Mr}}.\:\boldsymbol{\mathrm{Levent}} \\ $$

Question Number 165442 Answers: 1 Comments: 0

Question Number 165441 Answers: 1 Comments: 0

Question Number 165435 Answers: 1 Comments: 0

Question Number 165433 Answers: 1 Comments: 0

Question Number 165431 Answers: 2 Comments: 0

Question Number 165428 Answers: 1 Comments: 0

Question Number 165424 Answers: 0 Comments: 1

Question Number 165422 Answers: 0 Comments: 0

Question Number 165426 Answers: 0 Comments: 1

prove that ζ (0 )= ((−1)/2)

$$ \\ $$$$\:\:\:\:{prove}\:\:{that} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\zeta\:\left(\mathrm{0}\:\right)=\:\frac{−\mathrm{1}}{\mathrm{2}}\:\:\: \\ $$$$ \\ $$

Question Number 165419 Answers: 1 Comments: 0

Question Number 165418 Answers: 0 Comments: 0

prove that : 1^∗ : Σ_(n=1) ^∞ (( ζ (2n )−1)/( 1+ n)) = (3/(2 )) − ln (π ) 2^( ∗∗) : Σ_(n=2) ^∞ (( (−1)^( n) ( ζ (n )−1 ))/(1 + n))=(3/2) +(γ/2) −((ln(8π))/2) 3^( ∗∗) : Σ_(n=1) ^∞ (( ζ (2n )−1)/(1+ 2n)) = (3/2) −((ln(4π))/2) −−−− m.n −−−−

$$ \\ $$$$\:\:\:\:\:{prove}\:\:{that}\:: \\ $$$$ \\ $$$$\:\:\:\mathrm{1}^{\ast} :\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\zeta\:\left(\mathrm{2}{n}\:\right)−\mathrm{1}}{\:\mathrm{1}+\:{n}}\:=\:\frac{\mathrm{3}}{\mathrm{2}\:}\:\:−\:\mathrm{ln}\:\left(\pi\:\right) \\ $$$$\:\:\:\mathrm{2}^{\:\ast\ast} :\:\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\:\left(−\mathrm{1}\right)^{\:{n}} \left(\:\:\zeta\:\left({n}\:\right)−\mathrm{1}\:\right)}{\mathrm{1}\:+\:{n}}=\frac{\mathrm{3}}{\mathrm{2}}\:+\frac{\gamma}{\mathrm{2}}\:−\frac{\mathrm{ln}\left(\mathrm{8}\pi\right)}{\mathrm{2}} \\ $$$$\:\:\:\mathrm{3}^{\:\ast\ast} \::\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\zeta\:\left(\mathrm{2}{n}\:\right)−\mathrm{1}}{\mathrm{1}+\:\mathrm{2}{n}}\:=\:\frac{\mathrm{3}}{\mathrm{2}}\:−\frac{\mathrm{ln}\left(\mathrm{4}\pi\right)}{\mathrm{2}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:−−−−\:{m}.{n}\:−−−− \\ $$$$ \\ $$

Question Number 165404 Answers: 0 Comments: 9

given a loan from a bank on a fixed interest rate 5% suppose i want to make monthly payments for 14months how do i calculate the amount i should pay monthly and how do i know how much i will pay back to the bank at the end of the 14months? please i need explanations.

$$\mathrm{given}\:\mathrm{a}\:\mathrm{loan}\:\mathrm{from}\:\mathrm{a}\:\mathrm{bank}\:\mathrm{on}\:\mathrm{a}\:\mathrm{fixed}\:\mathrm{interest}\:\mathrm{rate}\:\mathrm{5\%} \\ $$$$\mathrm{suppose}\:\mathrm{i}\:\mathrm{want}\:\mathrm{to}\:\mathrm{make}\:\mathrm{monthly}\:\mathrm{payments}\:\mathrm{for}\:\mathrm{14months} \\ $$$$\mathrm{how}\:\mathrm{do}\:\mathrm{i}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{amount}\:\mathrm{i}\:\mathrm{should}\:\mathrm{pay}\:\mathrm{monthly}\:\mathrm{and} \\ $$$$\mathrm{how}\:\mathrm{do}\:\mathrm{i}\:\mathrm{know}\:\mathrm{how}\:\mathrm{much}\:\mathrm{i}\:\mathrm{will}\:\mathrm{pay}\:\mathrm{back}\:\mathrm{to}\:\mathrm{the}\:\mathrm{bank}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{end}\:\mathrm{of}\:\mathrm{the}\:\mathrm{14months}? \\ $$$$\mathrm{please}\:\mathrm{i}\:\mathrm{need}\:\mathrm{explanations}. \\ $$

Question Number 165403 Answers: 0 Comments: 0

sec^2 1° + sec^2 2° + sec^2 3° + …+ sec^2 89° = ?

$$\mathrm{sec}^{\mathrm{2}} \mathrm{1}°\:+\:\mathrm{sec}^{\mathrm{2}} \:\mathrm{2}°\:+\:\mathrm{sec}^{\mathrm{2}} \:\mathrm{3}°\:+\:\ldots+\:\mathrm{sec}^{\mathrm{2}} \:\mathrm{89}°\:=\:? \\ $$

Question Number 165402 Answers: 0 Comments: 1

Question Number 165532 Answers: 1 Comments: 0

Question Number 165392 Answers: 2 Comments: 0

lim_(x→+∞) ((e^(1/x) −cos (1/x))/(1−(√(1−(1/x^2 ))))) lim_(x→a) ((x^x −a^a )/(x−a))

$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\frac{{e}^{\frac{\mathrm{1}}{{x}}} −\mathrm{cos}\:\frac{\mathrm{1}}{{x}}}{\mathrm{1}−\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}} \\ $$$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\frac{{x}^{{x}} −{a}^{{a}} }{{x}−{a}} \\ $$

Question Number 165389 Answers: 2 Comments: 1

Question Number 165385 Answers: 1 Comments: 1

Number of ways..n differrnt things be distributed in r identical boxes so as 1)empty box is allowed 2)empty box not allowed Number of ways...n identical things be distributed to r identical boxes so as 1)empty box allowed 2)empty box not allowed

$${Number}\:{of}\:\:{ways}..{n}\: \\ $$$${differrnt}\:\:{things}\:{be}\:{distributed} \\ $$$${in}\:{r}\:{identical}\:{boxes}\:{so}\:{as} \\ $$$$\left.\mathrm{1}\right){empty}\:{box}\:{is}\:{allowed} \\ $$$$\left.\mathrm{2}\right){empty}\:{box}\:{not}\:{allowed} \\ $$$${Number}\:{of}\:{ways}...{n}\:{identical} \\ $$$${things}\:{be}\:{distributed}\:{to}\:{r} \\ $$$${identical}\:{boxes}\:{so}\:{as} \\ $$$$\left.\mathrm{1}\right){empty}\:{box}\:{allowed} \\ $$$$\left.\mathrm{2}\right){empty}\:{box}\:{not}\:{allowed} \\ $$$$ \\ $$

Question Number 165383 Answers: 0 Comments: 0

Find the value of this expression (−1)(2^2 −1)+(((3^2 −2))/(2!)) − (((4^2 −3))/(3!)) + …+ (((2019^2 −2018))/(2018!))

$${Find}\:\:{the}\:\:{value}\:\:{of}\:\:{this}\:\:{expression} \\ $$$$\left(−\mathrm{1}\right)\left(\mathrm{2}^{\mathrm{2}} −\mathrm{1}\right)+\frac{\left(\mathrm{3}^{\mathrm{2}} −\mathrm{2}\right)}{\mathrm{2}!}\:−\:\frac{\left(\mathrm{4}^{\mathrm{2}} −\mathrm{3}\right)}{\mathrm{3}!}\:+\:\ldots+\:\frac{\left(\mathrm{2019}^{\mathrm{2}} −\mathrm{2018}\right)}{\mathrm{2018}!} \\ $$

Question Number 165382 Answers: 0 Comments: 1

Question Number 165381 Answers: 2 Comments: 1

Question Number 165380 Answers: 1 Comments: 0

∫_0 ^( (π/2)) (( x^( 3) )/(sin^( 2) (x)))dx=^? (3/8) (π^( 2) ln(4)−7ζ(3))

$$ \\ $$$$\:\:\:\:\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{\:{x}^{\:\mathrm{3}} }{{sin}^{\:\mathrm{2}} \left({x}\right)}{dx}\overset{?} {=}\:\frac{\mathrm{3}}{\mathrm{8}}\:\left(\pi^{\:\mathrm{2}} {ln}\left(\mathrm{4}\right)−\mathrm{7}\zeta\left(\mathrm{3}\right)\right) \\ $$

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