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Question Number 105130 Answers: 0 Comments: 0

To now the real sequence in the follwing image : a_1 =1 ,a_2 =2 a_(nk +1) = ((a_(2k−1) +a_(2k) )/2) ∀ k ∈ Z^+ a_(2k+2) = (√(a_(2k) a_(2k+1) )) then prove that : lim_(n→∞) a_n = ((3(√3))/π)

$${To}\:{now}\:{the}\:{real}\:{sequence}\:{in}\:{the} \\ $$$${follwing}\:{image}\:: \\ $$$${a}_{\mathrm{1}} =\mathrm{1}\:\:\:,{a}_{\mathrm{2}} =\mathrm{2} \\ $$$${a}_{{nk}\:+\mathrm{1}} \:=\:\frac{{a}_{\mathrm{2}{k}−\mathrm{1}} \:+{a}_{\mathrm{2}{k}} }{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\forall\:{k}\:\in\:{Z}^{+} \\ $$$${a}_{\mathrm{2}{k}+\mathrm{2}} \:=\:\sqrt{{a}_{\mathrm{2}{k}} \:{a}_{\mathrm{2}{k}+\mathrm{1}} } \\ $$$${then}\: \\ $$$${prove}\:{that}\::\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{n}} \:=\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\pi} \\ $$$$ \\ $$

Question Number 105080 Answers: 2 Comments: 0

solve: x((x^2 −1)!) = 5((x−1)!)

$${solve}: \\ $$$$\:{x}\left(\left({x}^{\mathrm{2}} −\mathrm{1}\right)!\right)\:=\:\mathrm{5}\left(\left({x}−\mathrm{1}\right)!\right) \\ $$

Question Number 105075 Answers: 1 Comments: 0

(7/?) = ((21)/(27)) = ((7/3)/?)

$$\frac{\mathrm{7}}{?}\:=\:\frac{\mathrm{21}}{\mathrm{27}}\:=\:\frac{\frac{\mathrm{7}}{\mathrm{3}}}{?} \\ $$

Question Number 104975 Answers: 0 Comments: 2

Dear Forum-Friends There′s a trend to make answers short in the forum. To follow this trend some friends are omitting key-steps and for this reason the answers become difficult to understand and many readers like me are compelled to leave out such answers at the price of no-understanding!!! So please omit only calculation steps in order to make your answers short-&-understandable.

$$\mathrm{Dear}\:\mathrm{Forum}-\mathrm{Friends} \\ $$$$\:\:\:\:\:\:\mathcal{T}{here}'{s}\:{a}\:{trend}\:{to}\:{make} \\ $$$${answers}\:{short}\:{in}\:{the}\:{forum}. \\ $$$$\:\:\:\:\:\:\:\mathcal{T}{o}\:{follow}\:{this}\:{trend}\:{some} \\ $$$${friends}\:{are}\:{omitting}\:\boldsymbol{{key}}-\boldsymbol{{steps}} \\ $$$${and}\:{for}\:{this}\:{reason}\:{the}\:{answers} \\ $$$${become}\:\boldsymbol{{difficult}}\:\boldsymbol{{to}} \\ $$$$\boldsymbol{{understand}}\:{and}\: \\ $$$${many}\:{readers}\:{like}\:{me} \\ $$$$\:{are}\:{compelled}\:{to}\:{leave}\:{out} \\ $$$${such}\:{answers}\:{at}\:{the}\:{price}\:{of} \\ $$$$\boldsymbol{{no}}-\boldsymbol{{understanding}}!!! \\ $$$$\mathcal{S}{o}\:{please}\:{omit}\:{only} \\ $$$$\boldsymbol{{calculation}}\:\boldsymbol{{steps}}\:{in}\:{order}\:{to} \\ $$$${make}\:{your}\:{answers} \\ $$$${short}-\&-{understandable}. \\ $$

Question Number 104948 Answers: 0 Comments: 0

Question Number 104940 Answers: 1 Comments: 1

Question Number 104789 Answers: 0 Comments: 1

if y^((n)) is the derivative of the function y of the order n, then ∫y^((n)) dx =........

$${if}\:{y}^{\left({n}\right)} \:{is}\:{the}\:{derivative}\:{of}\:{the}\:{function}\:{y} \\ $$$${of}\:{the}\:{order}\:{n},\:{then} \\ $$$$\int{y}^{\left({n}\right)} {dx}\:=........ \\ $$

Question Number 104783 Answers: 2 Comments: 0

find (d^n y/dx^n ) for f(x)^ =(1/(√(1−x)))

$${find}\:\frac{{d}^{{n}} {y}}{{dx}^{{n}} }\:{for}\:\:\:\:{f}\left({x}\overset{} {\right)}=\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{x}}} \\ $$

Question Number 104777 Answers: 3 Comments: 0

Question Number 104768 Answers: 1 Comments: 0

((1/8)÷(1/8))((1/7)÷(1/7))((2/3)÷(2/3))= ?

$$\left(\frac{\mathrm{1}}{\mathrm{8}}\boldsymbol{\div}\frac{\mathrm{1}}{\mathrm{8}}\right)\left(\frac{\mathrm{1}}{\mathrm{7}}\boldsymbol{\div}\frac{\mathrm{1}}{\mathrm{7}}\right)\left(\frac{\mathrm{2}}{\mathrm{3}}\boldsymbol{\div}\frac{\mathrm{2}}{\mathrm{3}}\right)=\:? \\ $$

Question Number 104692 Answers: 2 Comments: 0

Question Number 104533 Answers: 3 Comments: 0

find : lim_(x→1) (((x+2)^2 +(x+1)^3 −17)/((√(x+3))−((x+7))^(1/3) ))

$${find}\:: \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\left({x}+\mathrm{2}\right)^{\mathrm{2}} +\left({x}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{17}}{\sqrt{{x}+\mathrm{3}}−\sqrt[{\mathrm{3}}]{{x}+\mathrm{7}}} \\ $$

Question Number 104494 Answers: 0 Comments: 0

Question Number 104471 Answers: 1 Comments: 0

Σ_(n=1) ^∞ ((e^n n!)/n^n )

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{e}^{{n}} {n}!}{{n}^{{n}} } \\ $$

Question Number 104406 Answers: 1 Comments: 1

Evaluate (1/2) + (3/8) + (3/(16)) + (5/(64)) + .....

$$\:\:\:\:\mathrm{Evaluate} \\ $$$$\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{3}}{\mathrm{8}}\:+\:\frac{\mathrm{3}}{\mathrm{16}}\:+\:\frac{\mathrm{5}}{\mathrm{64}}\:+\:..... \\ $$

Question Number 104296 Answers: 0 Comments: 0

FUN TIME AGAIN! S_n =1+2+3+4+5+6+7+8+9+.. S_n =1+(2+3+4)+(5+6+7)+... S_n =1+9+18+27+... S_n =1+9(1+2+3+4+5+6+7.......) S_n =1+9S_n S_n =−(1/8) S_n =1−1+1−1+1−1+1−1+1−1+.. S=(1/2) S_n =1−2+4−8+16−32+..... S_n =(1/(1+2))=(1/3) S_n =1+2+4+8+16+... S_n =1+2(1+2+4+8+...) S_n =1+2(1+2(1+2+4+8+...) S_n =1+2(1+2S_n ) −3S_n =3⇒S_n =−1

Question Number 104207 Answers: 2 Comments: 0

Question Number 104199 Answers: 2 Comments: 0

solve for real values of x the equation 4(3^(2x+1) )+17(3^x )=7. if m and n are positive real numbers other than 1, show that the log_n m+log_(1/m) n=0

$${solve}\:{for}\:{real}\:{values}\:{of}\:{x}\:{the}\:{equation} \\ $$$$\mathrm{4}\left(\mathrm{3}^{\mathrm{2}{x}+\mathrm{1}} \right)+\mathrm{17}\left(\mathrm{3}^{{x}} \right)=\mathrm{7}. \\ $$$${if}\:{m}\:{and}\:{n}\:{are}\:{positive}\:{real}\:{numbers}\:{other} \\ $$$${than}\:\mathrm{1},\:{show}\:{that}\:{the}\:\mathrm{log}_{{n}} {m}+\mathrm{log}_{\frac{\mathrm{1}}{{m}}} {n}=\mathrm{0} \\ $$

Question Number 104174 Answers: 2 Comments: 0

Π_(n=1) ^∞ ((n/(n+1)))^2

$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\frac{\mathrm{n}}{\mathrm{n}+\mathrm{1}}\right)^{\mathrm{2}} \\ $$

Question Number 104173 Answers: 0 Comments: 0

For fun! S_n =1+2+3+4+5+6+7+... 2S_n = 2 + 4 + 6 +.. −S_n =1+3+5+7+9+11+..... −(−(1/(12)))=1+3+5+7+9+11+... 1+3+5+7+9+....=(1/(12)) S_n =1+2+3+4+5+6+7+.... S_n =1+(1+1)+(1+2)+(1+3)+... S_n =(1+1+1+....)+S_n 1+1+1+1+1+1+....=0

$$\:\mathrm{For}\:\mathrm{fun}! \\ $$$$\mathrm{S}_{\mathrm{n}} =\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+\mathrm{7}+... \\ $$$$\mathrm{2S}_{\mathrm{n}} =\:\:\:\:\mathrm{2}\:\:+\:\:\:\mathrm{4}\:\:\:\:\:\:+\:\:\mathrm{6}\:\:\:\:\:\:+.. \\ $$$$−\mathrm{S}_{\mathrm{n}} =\mathrm{1}+\mathrm{3}+\mathrm{5}+\mathrm{7}+\mathrm{9}+\mathrm{11}+..... \\ $$$$−\left(−\frac{\mathrm{1}}{\mathrm{12}}\right)=\mathrm{1}+\mathrm{3}+\mathrm{5}+\mathrm{7}+\mathrm{9}+\mathrm{11}+... \\ $$$$\mathrm{1}+\mathrm{3}+\mathrm{5}+\mathrm{7}+\mathrm{9}+....=\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$ \\ $$$$\mathrm{S}_{\mathrm{n}} =\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+\mathrm{7}+.... \\ $$$$\mathrm{S}_{\mathrm{n}} =\mathrm{1}+\left(\mathrm{1}+\mathrm{1}\right)+\left(\mathrm{1}+\mathrm{2}\right)+\left(\mathrm{1}+\mathrm{3}\right)+... \\ $$$$\mathrm{S}_{\mathrm{n}} =\left(\mathrm{1}+\mathrm{1}+\mathrm{1}+....\right)+\mathrm{S}_{\mathrm{n}} \\ $$$$\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+....=\mathrm{0} \\ $$

Question Number 104333 Answers: 2 Comments: 0

if i^( 5n +1 ) = 1 , n∈Z then show that : n = 3+4p ,p∈Z

$${if}\:{i}^{\:\mathrm{5}{n}\:+\mathrm{1}\:} =\:\mathrm{1}\:\:\:\:,\:{n}\in{Z} \\ $$$${then}\:{show}\:{that}\::\:{n}\:=\:\mathrm{3}+\mathrm{4}{p}\:\:\:,{p}\in{Z} \\ $$

Question Number 104176 Answers: 1 Comments: 0

Σ_(n=1) ^∞ (((n/(n+1)))^2 −(2/(n+1))−1)

$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\left(\frac{\mathrm{n}}{\mathrm{n}+\mathrm{1}}\right)^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{n}+\mathrm{1}}−\mathrm{1}\right) \\ $$

Question Number 103808 Answers: 0 Comments: 1

muhun kitu pisan

$${muhun}\:{kitu}\:{pisan} \\ $$

Question Number 103669 Answers: 2 Comments: 0

prove that : a) ∫_(−3) ^(−1) x^2 dx ≥∫_1 ^3 (2x−1)dx b)∫_(−2) ^0 xdx ≤∫_0 ^2 (x^2 + x )dx c)∫_1 ^4 (x^2 + 2)dx ≥∫_2 ^5 (2x −5)dx d)∫_(−π) ^(−((3π)/4)) cos 2x dx ≥∫_((3π)/4) ^π sin 2x dx

$${prove}\:{that}\:: \\ $$$$\left.{a}\right)\:\int_{−\mathrm{3}} ^{−\mathrm{1}} {x}^{\mathrm{2}} {dx}\:\geqslant\int_{\mathrm{1}} ^{\mathrm{3}} \left(\mathrm{2}{x}−\mathrm{1}\right){dx} \\ $$$$\left.{b}\right)\int_{−\mathrm{2}} ^{\mathrm{0}} {xdx}\:\leqslant\int_{\mathrm{0}} ^{\mathrm{2}} \left({x}^{\mathrm{2}} \:+\:{x}\:\right){dx} \\ $$$$\left.{c}\right)\int_{\mathrm{1}} ^{\mathrm{4}} \left({x}^{\mathrm{2}} \:+\:\mathrm{2}\right){dx}\:\:\geqslant\int_{\mathrm{2}} ^{\mathrm{5}} \left(\mathrm{2}{x}\:−\mathrm{5}\right){dx} \\ $$$$\left.{d}\right)\int_{−\pi} ^{−\frac{\mathrm{3}\pi}{\mathrm{4}}} \mathrm{cos}\:\mathrm{2}{x}\:{dx}\:\geqslant\int_{\frac{\mathrm{3}\pi}{\mathrm{4}}} ^{\pi} \mathrm{sin}\:\mathrm{2}{x}\:{dx} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 103574 Answers: 0 Comments: 0

Question Number 103510 Answers: 1 Comments: 5

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