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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 =........

$i f y^{(n)} i s t h e d e r i v a t i v e o f t h e f u n c t i o n y$ $o f t h e o r d e r n, t h e n$ $\int y^{(n)} d x = . . . . . . . .$

Question Number 104783 Answers: 2 Comments: 0

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

$Missing \left or extra \right$

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))= ?

$(\frac{1}{8} \div \frac{1}{8}) (\frac{1}{7} \div \frac{1}{7}) (\frac{2}{3} \div \frac{2}{3}) = ?$

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) ))

$f i n d :$ $\lim_{x \to 1} \frac{{(x + 2)}^{2} + {(x + 1)}^{3} - 17}{\sqrt{x + 3} - \sqrt[3]{x + 7}}$

Question Number 104494 Answers: 0 Comments: 0

Question Number 104471 Answers: 1 Comments: 0

Σ_(n=1) ^∞ ((e^n n!)/n^n )

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

Question Number 104406 Answers: 1 Comments: 1

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

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

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

$FUN TIME AGAIN!$ $S_{n} = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + . .$ $S_{n} = 1 + (2 + 3 + 4) + (5 + 6 + 7) + . . .$ $S_{n} = 1 + 9 + 18 + 27 + . . .$ $S_{n} = 1 + 9 (1 + 2 + 3 + 4 + 5 + 6 + 7 . . . . . . .)$ $S_{n} = 1 + {9 S}_{n}$ $S_{n} = - \frac{1}{8}$ $S_{n} = 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + . .$ $S = \frac{1}{2}$ $S_{n} = 1 - 2 + 4 - 8 + 16 - 32 + . . . . .$ $S_{n} = \frac{1}{1 + 2} = \frac{1}{3}$ $S_{n} = 1 + 2 + 4 + 8 + 16 + . . .$ $S_{n} = 1 + 2 (1 + 2 + 4 + 8 + . . .)$ $S_{n} = 1 + 2 (1 + 2 (1 + 2 + 4 + 8 + . . .)$ $S_{n} = 1 + 2 (1 + {2 S}_{n})$ $- {3 S}_{n} = 3 \Rightarrow S_{n} = - 1$

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

$s o l v e f o r r e a l v a l u e s o f x t h e e q u a t i o n$ $4 (3^{2 x + 1}) + 17 (3^{x}) = 7 .$ $i f m a n d n a r e p o s i t i v e r e a l n u m b e r s o t h e r$ $t h a n 1, s h o w t h a t t h e \log_{n} m + \log_{\frac{1}{m}} n = 0$

Question Number 104174 Answers: 2 Comments: 0

Π_(n=1) ^∞ ((n/(n+1)))^2

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

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

$For fun!$ $S_{n} = 1 + 2 + 3 + 4 + 5 + 6 + 7 + . . .$ ${2 S}_{n} = 2 + 4 + 6 + . .$ $- S_{n} = 1 + 3 + 5 + 7 + 9 + 11 + . . . . .$ $- (- \frac{1}{12}) = 1 + 3 + 5 + 7 + 9 + 11 + . . .$ $1 + 3 + 5 + 7 + 9 + . . . . = \frac{1}{12}$ $S_{n} = 1 + 2 + 3 + 4 + 5 + 6 + 7 + . . . .$ $S_{n} = 1 + (1 + 1) + (1 + 2) + (1 + 3) + . . .$ $S_{n} = (1 + 1 + 1 + . . . .) + S_{n}$ $1 + 1 + 1 + 1 + 1 + 1 + . . . . = 0$

Question Number 104333 Answers: 2 Comments: 0

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

$i f i^{5 n + 1} = 1, n \in Z$ $t h e n s h o w t h a t : n = 3 + 4 p, p \in Z$

Question Number 104176 Answers: 1 Comments: 0

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

$\underset{n = 1}{\sum^{\infty}} ({(\frac{n}{n + 1})}^{2} - \frac{2}{n + 1} - 1)$

Question Number 103808 Answers: 0 Comments: 1

muhun kitu pisan

$m u h u n k i t u p i s a n$

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

$p r o v e t h a t :$ $a) \int_{- 3}^{- 1} x^{2} d x ⩾ \int_{1}^{3} (2 x - 1) d x$ $b) \int_{- 2}^{0} x d x ⩽ \int_{0}^{2} (x^{2} + x) d x$ $c) \int_{1}^{4} (x^{2} + 2) d x ⩾ \int_{2}^{5} (2 x - 5) d x$ $d) \int_{- π}^{- \frac{3 π}{4}} \cos 2 x d x ⩾ \int_{\frac{3 π}{4}}^{π} \sin 2 x d x$

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

$S o l v e :$ $3^{x} = 4 x$

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

$1 + 1 + 1 + 1 + 1 + 1 + 1 + . . . . = S_{n}$ $S_{n} = 1 + 1 + 1 + 1 + 1 + 1 + 1 + . . . .$ $2 S_{n} = 2 + 2 + 2 + . . . . . . .$ $. . . . . . . . . . s u b t r a c t i n g$ $- S_{n} = 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + . . . .$ $- S_{n} = \frac{1}{2}$ $S_{n} = - \frac{1}{2}$ $I h a v e f o u n d t h i s w h i l e e x p e r i m e n t . I k n o w t h e s u m d i v e r g e s$ $b u t i s i t p r e t t y c o o l ?$ $K i n d l y r e c t i f y m e i f t h e r e i s a n y f a u l t o n t h i s n o n r i g o r o u s$ $p r o c e s s$ $I h a v e f o u n d s o m e R a m a n u j a n p r o o f$ $S_{n} = 1 + 2 + 3 + 4 + 5 + 6 + 7 + . . .$ $4 S_{n} = 4 + 8 + 12 + . . .$ $- 3 S_{n} = 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + . . . . . .$ $- 3 S_{n} = \frac{1}{4}$ $S_{n} = - \frac{1}{12}$ $R a m a n u j a n h a d d o n e t h i s o n h i s n o t e b o o k$

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