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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

Question Number 103504 Answers: 0 Comments: 0

Question Number 103503 Answers: 2 Comments: 0

Solve : 3^x = 4x

$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:{Solve}\:: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{3}^{{x}} \:=\:\mathrm{4}{x} \\ $$

Question Number 103236 Answers: 0 Comments: 3

e^e^(e.....∞) =?

$${e}^{{e}^{{e}.....\infty} } =? \\ $$

Question Number 103219 Answers: 1 Comments: 3

1+1+1+1+1+1+1+....=S_n S_n =1+1+1+1+1+1+1+.... 2S_n = 2 + 2 + 2+....... .......... subtracting −S_n =1−1+1−1+1−1+1−1+1−1+.... −S_n =(1/2) S_n =−(1/2) I have found this while experiment . I know the sum diverges but is it pretty cool? Kindly rectify me if there is any fault on this non rigorous process I have found some Ramanujan proof S_n =1+2+3+4+5+6+7+... 4S_n = 4+ 8 + 12+... −3S_n =1−2+3−4+5−6+7−8+...... −3S_n =(1/4) S_n =−(1/(12)) Ramanujan had done this on his notebook

Question Number 103209 Answers: 3 Comments: 0

Question Number 103119 Answers: 0 Comments: 0

∫((log(((1+(√5))/2)(√x)−1))/(x^(√x) log(((1+(√5))/2)(√x)+1)−1))

$$\int\frac{{log}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\sqrt{{x}}−\mathrm{1}\right)}{{x}^{\sqrt{{x}}} {log}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\sqrt{{x}}+\mathrm{1}\right)−\mathrm{1}} \\ $$

Question Number 103016 Answers: 0 Comments: 0

Question Number 103859 Answers: 1 Comments: 0

solve : y^2 + x^2 = −sin(θ) ., (π/2)≥ θ≥−(π/2)

$$\:\:\: \\ $$$$\:\:\:\:\:\:{solve}\:: \\ $$$$\:\:\:\:\:\:{y}^{\mathrm{2}} \:+\:{x}^{\mathrm{2}} \:=\:−\mathrm{sin}\left(\theta\right)\:\:\:\:\:.,\:\:\:\:\frac{\pi}{\mathrm{2}}\geqslant\:\theta\geqslant−\frac{\pi}{\mathrm{2}} \\ $$

Question Number 102822 Answers: 1 Comments: 1

Σ_(n=1) ^∞ ((n!)/n^n )

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

Question Number 102745 Answers: 0 Comments: 2

1−1+1−1+1−1+1−1+.....=(1/2) {But it diverges 1+1+1+1+1+1+1+......=−(1/2) {But it diverges 1+2+4+8+16+.....=−1 {But it diverges 1+2+3+4+5+6+7=−(1/(12)) {But it diverges 1−2+4−8+.....=(1/3) {But it diverges 1−2+3−4+5−6+.....=(1/4) {Is it a divergent?????

$$\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+.....=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\left\{{But}\:{it}\:{diverges}\right. \\ $$$$\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+......=−\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\left\{{But}\:{it}\:{diverges}\right. \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{4}+\mathrm{8}+\mathrm{16}+.....=−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{{But}\:{it}\:{diverges}\right. \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+\mathrm{7}=−\frac{\mathrm{1}}{\mathrm{12}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{{But}\:{it}\:{diverges}\right. \\ $$$$\mathrm{1}−\mathrm{2}+\mathrm{4}−\mathrm{8}+.....=\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{{But}\:{it}\:{diverges}\right. \\ $$$$\mathrm{1}−\mathrm{2}+\mathrm{3}−\mathrm{4}+\mathrm{5}−\mathrm{6}+.....=\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\:\:\left\{{Is}\:{it}\:{a}\:{divergent}?????\right. \\ $$

Question Number 102701 Answers: 2 Comments: 1

Evaluate: ∫((sin x)/x)dx

$${Evaluate}: \\ $$$$\int\frac{\mathrm{sin}\:{x}}{{x}}{dx} \\ $$

Question Number 102686 Answers: 0 Comments: 1

Σ_(r=1) ^∞ i^r +Σ_(r=0) ^∞ i^r

$$\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}{i}^{{r}} +\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}{i}^{{r}} \\ $$

Question Number 102664 Answers: 2 Comments: 0

2x=5 x=? −−− for test app only

$$\mathrm{2}{x}=\mathrm{5} \\ $$$${x}=? \\ $$$$−−− \\ $$$$\boldsymbol{{for}}\:\boldsymbol{{test}}\:\boldsymbol{{app}}\:\boldsymbol{{only}} \\ $$

Question Number 102651 Answers: 0 Comments: 1

12+14+24+58+164....upto nth terms

$$\mathrm{12}+\mathrm{14}+\mathrm{24}+\mathrm{58}+\mathrm{164}....\mathrm{upto}\:\mathrm{nth}\:\mathrm{terms} \\ $$

Question Number 102635 Answers: 2 Comments: 0

Question Number 102627 Answers: 3 Comments: 0

show that: cosθ + cos2θ + ....cos nθ= ((cos (1/2)(n +1)θ sin(1/2)nθ)/(sin (1/2)nθ)) Show that: sin θ + sin 2θ + ....+ sin nθ = ((sin (1/2)(n + 1)θ sin(1/2)nθ)/(sin (1/2)nθ)) where θ ∈ R and θ ≠2πk , k ∈Z

$$\mathrm{show}\:\mathrm{that}:\:\mathrm{cos}\theta\:+\:\mathrm{cos2}\theta\:+\:....\mathrm{cos}\:{n}\theta=\:\frac{\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}\left({n}\:+\mathrm{1}\right)\theta\:\mathrm{sin}\frac{\mathrm{1}}{\mathrm{2}}{n}\theta}{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{n}\theta} \\ $$$$\mathrm{Show}\:\mathrm{that}:\:\mathrm{sin}\:\theta\:+\:\mathrm{sin}\:\mathrm{2}\theta\:+\:....+\:\mathrm{sin}\:{n}\theta\:=\:\frac{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\left({n}\:+\:\mathrm{1}\right)\theta\:\mathrm{sin}\frac{\mathrm{1}}{\mathrm{2}}{n}\theta}{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{n}\theta} \\ $$$$\mathrm{where}\:\theta\:\in\:\mathbb{R}\:\mathrm{and}\:\theta\:\neq\mathrm{2}\pi{k}\:,\:{k}\:\in\mathbb{Z} \\ $$$$ \\ $$

Question Number 102598 Answers: 2 Comments: 0

Question Number 102545 Answers: 1 Comments: 0

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