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The numerator of the first term above should look like this:  

\(\ln[(1+1/n)^n (n+1)]\)

i.e. all the terms containing n are part of the argument of the logarithm function.

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0 mins ago

The price of a scooter is € 2500.

Henk has not saved enough money to pay for the scooter.

He borrows € 1500 from his parents.

How many Euros has Henk for the scooter ?

Henk has € 2500 - € 1500 = € 1000

Like so:

\(f(n)=\frac{\ln(1+1/n)^n(n+1)}{\ln(n^n)\ln(n+1)^{n+1}}\\\\ f(n)=\frac{(n+1)\ln(n+1)-n\ln(n)}{n\ln(n)(n+1)\ln(n+1)}\\\\ f(n)=\frac{1}{n\ln(n)}-\frac{1}{(n+1)\ln(n+1)}\\\\ \Sigma_{n=2}^\infty f(n)=\frac{1}{2\ln2}-\frac{1}{3ln3}+\frac{1}{3ln3}-\frac{1}{4ln4}+\frac{1}{5ln5}-\frac{1}{5ln5}+...\\\\ \Sigma_{n=2}^\infty f(n)=\frac{1}{2ln2}\)

i.e. the only term that doesn't get cancelled out is the first one.

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I think the translation leaves something to be desired. The meaning has been lost to me.  

Where be thine castle ?

I talked about this yesterday but my post disappeared. :(

I think that guests Wolfram Alpha input is incorrect.  

I have just copied and pasted guests input into Wolfram|Alpha and it was not entered correctly and Wolfram Alpha said it was equal to infinity...

Is this what you want?

\( (4y-\frac{5 }{2}) - (2y-\frac{1}{6})\)

what is the area of  A(0,3) B(4,-1) C(-2,1) D(2,-3)

\(\begin{array}{|rcll|} \hline \text{Let } \vec{A} &=& \binom{0}{3} \\ \text{Let } \vec{B} &=& \binom{4}{-1} \\ \text{Let } \vec{C} &=& \binom{-2}{1} \\ \text{Let } \vec{D} &=& \binom{2}{-3} \\ \hline \end{array}\)

Points clockwise:

\(\begin{array}{|rcll|} \hline 1. & \vec{A} &=& \binom{0}{3} \\ 2. & \vec{B} &=& \binom{4}{-1} \\ 3. & \vec{D} &=& \binom{2}{-3} \\ 4. & \vec{C} &=& \binom{-2}{1} \\ \hline \end{array}\)

Area:

\(\begin{array}{|rcll|} \hline 2A &=& |\vec{B} \times \vec{A}| + |\vec{D} \times \vec{B}| + |\vec{C} \times \vec{D}| + |\vec{A} \times \vec{C}| \\ 2A &=& |\binom{4}{-1} \times \binom{0}{3}| + |\binom{2}{-3} \times \binom{4}{-1}| + |\binom{-2}{1} \times \binom{2}{-3}| + |\binom{0}{3} \times \binom{-2}{1}| \\ 2A &=& 4\cdot 3-(-1)\cdot 0 +2(-1)-(-3)\cdot 4 +(-2)(-3)-1\cdot2 + 0\cdot 1 -3(-2) \\ 2A &=& 12-0 -2+12 +6-2 + 0 +6 \\ 2A &=& 32 \\ \mathbf{A} & \mathbf{=}& \mathbf{16} \\ \hline \end{array}\)

.....but I didn't get all of the possible answers.....just like I didn't finish that last post before hitting 'enter'     LOL  

Thanx for the heads up Chris.....I did check the results to be sure before I posted (this time)  !

Hi EP,

Don't be concerned, 

THIS bear did NOT eat THIS dog.         

sinx-2sinxcosx=0

Formula:

\(\begin{array}{|rcll|} \hline 2\ sin(x) \cos(x) &=& \sin(2x)\\ \hline \end{array}\)

So:

\(\begin{array}{|rcll|} \hline 0 &=& \sin(x) - 2\cdot sin(x) \cos(x) \quad & | \quad 2\cdot sin(x) \cos(x) = \sin(2x) \\ 0 &=& \sin(x) - \sin(2x) \quad & | \quad + \sin(2x) \\ \sin(2x) &=& \sin(x) \\ \hline \end{array} \)

First result:

\(\begin{array}{|rcll|} \hline \sin(2x) &=& \sin(x) \quad & | \quad \arcsin() \text{ both sides } \\ 2x &=& x \quad & | \quad -x \\ 2x-x &=& 0 \\ \mathbf{ x } & \mathbf{=} & \mathbf{ 0 \pm 360^{\circ} \cdot n } \quad & | \quad n \in N \\ \hline \end{array}\)

Second result:

\(\begin{array}{|rcll|} \hline \sin(2x) &=& \sin(x) \quad & | \quad \sin(x) = \sin(180^{\circ}-x) \\ \sin(2x) &=& \sin(180^{\circ}-x) \quad & | \quad \arcsin() \text{ both sides } \\ 2x &=& 180^{\circ}-x \quad & | \quad +x \\ 2x+x &=& 180^{\circ} \\ 3x &=& 180^{\circ} \quad & | \quad : 3 \\ x &=& 60^{\circ} \\ \mathbf{ x } & \mathbf{=} & \mathbf{ 60^{\circ} \pm 360^{\circ} \cdot n } \quad & | \quad n \in N \\ \hline \end{array}\)

Third result:

\(\begin{array}{|rcll|} \hline \sin(2x) &=& \sin(x) \quad & | \quad -\sin(x) = \sin(180^{\circ}+x) \\ \sin(2x) &=& -\sin(180^{\circ}+x) \\ \sin(2x) &=& \sin(-180^{\circ}-x) \quad & | \quad \arcsin() \text{ both sides } \\ 2x &=& -180^{\circ}-x \quad & | \quad +x \\ 2x+x &=& -180^{\circ} \\ 3x &=& -180^{\circ} \quad & | \quad : 3 \\ x &=& -60^{\circ} \\ \mathbf{ x } & \mathbf{=} & \mathbf{ -60^{\circ} \pm 360^{\circ} \cdot n } \quad & | \quad n \in N \\ \hline \end{array} \)

6.  -<4~/=<1, <3~/=<2

     -AA similarity postulate

     -BA/CB=DE/CD

Wait... You speak Russian? Or was that google translate?

There is no button like there is on many calculators

Там нет кнопки, как есть на многих калькуляторах

if it is   27.92535 degrees then that becomes  

\(27.92535^\circ\\=27^\circ+0.92535^\circ\\ =27^\circ+0.92535*60'\\ =27^\circ+55.521'\\ =27^\circ+55'+0.521*60''\\ =27^\circ+55'+31.26''\\ =27^\circ\; 55' \;31.26''\\\)

2. -Def. of angle bisector

5. -<ADB~/=<CDF

Unsure about 9

Here is the visual for #6

I'm not sure, but you can find interesting information on this site...http://physics.stackexchange.com/questions/26696/how-do-we-determine-the-mass-of-a-black-hole

5. Angle-Angle Similarity Postulate

    proportional 

    segment addition postulate

6. Need a diagram because I'm unsure about the positioning  of angles 1,2,3, and 4

Yes, that is correct haha

Solved it!

To shorten 2048 /2 /2 /2 /2 /2 /2 /2 /2

it would be 2048/2^8

Number/divideBy^#ofTimesToDivide

Oh, but that was an example. I was just wondering if there was an equation to shorten it.

Actually, this MIGHT work right?

(2048/2)8

Would that represent 2048/2/2/2/2/2/2/2/2 correctly?

Yes, but it's a great question, so don't feel bad. :) The answer to your question is yes. 

I'm not sure about the multiplication part, you would need to give a number times a number, and see if it works or not. But 2048/2/2/2/2/2/2/2/2 =8. And the shortest way you can figure that out is to also do 2048/2^8. Your welcome :)