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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

Question Number 103504 Answers: 0 Comments: 0

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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

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