All Answers 358098 Answers

I didn't even notice thanks i will use \\\\ and i will put in \sqrt

$$\\\\\left(z-\frac{2}{\sqrt{z}}\right)\\\\The\:general\:term\:is\\\\
=(9Cr)(z)(9-r)\left(\frac{z^9}{z^r}\right)\left(\frac{(-2)^r}{(\sqrt{z}^r}\right)\\\\
=(9Cr)\left(\frac{z^9*(-2)^r}{z^r*(\sqrt{z}^r)}\right)\\\\
=(9Cr)\left(\frac{*(-2)^r}{z^(2r/2)*(z)^(r/2)}\right)\\\\
=(9Cr)\left(\frac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\\\
=(9Cr)(z^{((18-3r)/2)}*(-2)^r)
\\\\The\:constant\:term\:will\:be\:when\\\\
(18-3r)/2=0\\18-3r=0\\r=6$$

What to do next.....Hmmmm...

I dont think my math knowledge has exceeded far enough to think of a new topic :)

Your LaTex abilities are improving all the time, MG1....!!!!

Maybe....one day......you can teach me....LOL  !!!!!

I agree!

Thank you Melody!  :D    :D     :D

aha Melody,  I've noticed Heureka's LaTeX ever sinced i layed my eyes on this website and LaTeX.

The LaTeX is amazing but i think i should take it slow before i get to your level...Really slow :D

Absolutely :)

// for next line //// for bigger space ////// for bigger again etc.

I like mine well spaced so usually I use ////  Occasionally i will use even more.  :)

The only thing that I can see wrong with that is one missing backslash in front of a sqrt.

Great job MG.  What do you want to do next?

Well....we couldn't calculate the sum of all of them....!!!  {the sum would be infinite !!!}

But...we have a "formula" to calculate the sum of the first N of them.....

Let's take a simple example to see if we can find out what it is.......

Notice that.......if we had an some number of integers  to add....we might have something like this.....

1 + 2 + 3 + .....+ N-2 + N-1 + N

Now, let's rewrite the same sum backwards and add both things......so we have

    [ 1              + 2             +     3 +.......         + N-2         + N-1           + N ]

+ [N              + N-1         +    N-2 +    .....      + 3             +  2             + 1]

= [N + 1]     + [N + 1]    + [N + 1] +  .....      +[N + 1]    +[N + 1]     +[ N + 1]

Now, notice that we have "N" pairs of  [N + 1] terms.....so the sum of these is just  ...  N(N+ 1)

But.......we've summed the series twice.....so we need to divide the above result by 2  

And we arrive at the "formula" for the sum of the first N integers......N(N + 1) / 2    .......  !!!!

I'm on Nauseated side...This Titanium army has gone too far. It's even changing this sites name.

This can not be defined because :

1. There's a contradiction: 0 divide by anything = 0 but anything divide by itself = 1;
2. You can't divide by 0

@@ End of Day Wrap   Sat 9/5/15   Sydney,   Australia Time    2:20am (Really Sun) ♪ ♫

Hi everyone,

I did not write a wrap yesterday so today is a big one.  We have so many wonderful answerers, and here they all are:  Martind, Cstaib45, Alan, Squaddy, DarkBlaze347, MathsGod1, HeadingNorth, CPhill, Nathan0501, Everymornin17, Awsomeee, SilkerR7, Civonamzuk, Zacismyname, Mihail, annabelle1DirectionS, Geno3141, Xerces, Fiora, EinsteinJr and MrGenius.  A big thank you to each of you.    

------------

I saved this post for the wrap because it was extremely funny.

I have whited out a number of offensive posts that it did contain.

Good job.

You both have earned a cookie:

Thanks Melody, MrGenius and MathsGod1;

There are two systems: The short system, or short scale, and the long system, or long scale.

In the short system:

1,000,000,000 is a billion

1,000,000,000,000 is a trillion,

1,000,000,000,000,000 is a quadrillion, et caetera...

In the long system:

1,000,000,000 is a thousand million or, more occasionally, a milliard (I love this word )

1,000,000,000,000 is a billion

1,000,000,000,000,000 is a thousand billion or a billiard

and so on.

The short system has always been used in USA; it was officially adopted by UK in 1974.

But the long system is still used in almost every non-English-speaking countries in the World.

Thanks, Melody....I really liked this one , too!!!!!

$$\\\left(z-\dfrac{2}{\sqrt{z}}\right)\\\\The\:general\:term\:is\\\\
=(9Cr)(z)(9-r)\left(\dfrac{z^9}{z^r}\right)\left(\dfrac{(-2)^r}{(\sqrt{z}^r}\right)\\
=(9Cr)\left(\dfrac{z^9*(-2)^r}{z^r*(sqrt{z}^r)}\right)\\
=(9Cr)\left(\dfrac{*(-2)^r}{z^(2r/2)*(z)^(r/2)}\right)\\
=(9Cr)\left(\dfrac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\
=(9Cr)(z^{((18-3r)/2)}*(-2)^r)
\\The\:constant\:term\:will\:be\:when\\
(18-3r)/2=0\\18-3r=0\\r=6$$

Is there a way to leave larger gaps it seems so cramped.

Sun 10/5/15

If you would like to comment on other site issues please do so on the Lantern Thread.  Thank you.    

Interest Posts:

FTJ means "For the Juniors"

thanks Nicholas for joining in :)

You can always put up a puzzle just make sure you know what a good answer is first :)

Very nice, heureka....!!!!

I'll try :))

I have never heard that term either.  Ketet             Thanks Squaddy

You are welcome MG

If you try those sites you can get back with a review of them.  That would be great.  :))

 $$$(3x+(2y+1))^2$$

This term would be

C(2,1)*(3x)^1 *(2y + 1)^1  =

2(3x)(2y + 1) =

6x(2y + 1) =

12xy + 6x

So....the coefficient is 12

Like WOW Heureka,    

MathsGod1 is learning LaTex.  He is only 12 but he is learning very quickly.

I should send him to you  :))

I'm quite happy to keep teaching him but you can if you would like to.

You are Lord of LaTex in Camelot. :))  

I am sorry but I do not understand the question. 

I gave you thumbs up for asking - I like everyone to ask questions.  I just don't understand  what you mean.