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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: lacoste on March 19, 2009, 08:13:29 pm

Title: binomial no calc
Post by: lacoste on March 19, 2009, 08:13:29 pm: how to do this by hand with no calc.( i dont know how to use pascals triangle fully, so could you please explain if you use it)

binomial expansion.

(2/x+3x)^5

thanks
Title: Re: binomial no calc
Post by: TrueTears on March 19, 2009, 08:16:07 pm: do you mean $(\frac{2}{x+3x})^5 or (\frac{2}{x} + 3x)^5$ ? lol
Title: Re: binomial no calc
Post by: lacoste on March 19, 2009, 08:17:04 pm: oh sorry truetears, the second one; where 2 is over x by itself only
Title: Re: binomial no calc
Post by: TrueTears on March 19, 2009, 08:42:09 pm: Alright nps

Consider this:

let $z = \frac{2}{x}$

$y = 3x$

$(y+z)^5$ Now consider y as having the 'x' and z is just a constant, because the expression in the brackets must have a linear expression to do binomial theorem.

use the binomial theorem on that yields...

$5C_0(y)^5(z)^0 + 5C_1(y)^4(z)^1 + 5C_2(y)^3(z)^2 + 5C_3(y)^2(z)^3 + 5C_4(y)^1(z)^4 + 5C_5 (y)^0(z)^5$

$y^5 + 5y^4z +10y^3z^2 + 10y^2z^3 + 5yz^4 +z^5$

Now sub in the values for y and z and there's your answer :)
Title: Re: binomial no calc
Post by: dekoyl on March 19, 2009, 08:43:49 pm: 1
11
121
1331
14641
1,5,10,5,1

$(1)\left(\frac{2}{x}\right)^{5}\left(3x\right)^{0} + (5)\left(\frac{2}{x}\right)^{4}\left(3x\right)^{1}....(1)\left(\frac{2}{x}\right)^{0}\left(3x\right)^{5}$

I haven't done this in a looong time :P Hope you understand. Is this right?
Title: Re: binomial no calc
Post by: TrueTears on March 19, 2009, 08:45:54 pm: yeah and dekoyl's way is also correct, he let z as having the 'x' and y as the constant. :)
Title: Re: binomial no calc
Post by: lacoste on March 19, 2009, 08:46:41 pm: oh yes thanks.

and

find the coefficient of x^2 in the above q

This is such a difficulty
Title: Re: binomial no calc
Post by: lacoste on March 19, 2009, 08:48:38 pm: without part a which was expanding it all
Title: Re: binomial no calc
Post by: TrueTears on March 19, 2009, 08:49:01 pm: There are no coefficients with $x^2$ o.o?
Title: Re: binomial no calc
Post by: lacoste on March 19, 2009, 08:49:39 pm: Quote from: TrueTears on March 19, 2009, 08:49:01 pm
There are no coefficients with $x^2$ o.o?

Yes, the ans is 0 but how do i know that its 0 without expanding it all out??
Title: Re: binomial no calc
Post by: dekoyl on March 19, 2009, 08:51:50 pm: Quote from: lacoste on March 19, 2009, 08:49:39 pm
Quote from: TrueTears on March 19, 2009, 08:49:01 pm
There are no coefficients with $x^2$ o.o?

Yes, the ans its 0 but how do i know that theres none with expanding it all out??
Just do a quick observation. :) No need to expand it out. You just have to cancel a few indicies in your head.
Title: Re: binomial no calc
Post by: lacoste on March 19, 2009, 08:53:31 pm: sorry dekoyl but how do i do that?

where?

im not very good
Title: Re: binomial no calc
Post by: TrueTears on March 19, 2009, 08:54:53 pm: You would do this again

let $y = 3x$ and $z = \frac{2}{x}$

now $(y+z)^5$

so, $x^2$ term means the 4th term from the highest right? (ie, $x^5, x^4, x^3, x^2$ )

so lets consider y as containing the 'x' and z as the constant

(remember the term that contains the 'x' always has the power of what you are looking for, so the power of 2 would mean $y^2$ , the power of 3 would mean $y^3$ etc etc)

the fourth term would be $5C_3(y)^2(z)^3 = \frac{720}{x}$ which does not contain a $x^2$

now lets consider z as containing the 'x' and y as the constant

the fourth term would be $5C_3(z)^2(y)^3 = 1080x$ again this does not contain a $x^2$

Therefore coefficient of $x^2 = 0$

Hope this helps.

EDIT: dekoyl is just too godly :)
Title: Re: binomial no calc
Post by: dekoyl on March 19, 2009, 08:58:30 pm: Quote from: lacoste on March 19, 2009, 08:53:31 pm
sorry dekoyl but how do i do that?

where?

im not very good
It's okay.

Well the whole thing is to the power of 5. So looking at the two "parts" inside the brackets ( $\frac{2}{x} , 3x$ ) you will see that the closest will either be (note k = a constant) $\frac{kx^{4}}{x^1} = kx^3$ or $\frac{kx^{3}}{x^2} = kx$ so it's 3 then 1. There is no $kx^{2}$ .

Bah I suck at explaining :P Hope you understand.

Edit: Wow TrueTears is a beast at typing and latex.
Title: Re: binomial no calc
Post by: lacoste on March 19, 2009, 09:05:15 pm: thankyou everyone!!!!!!

truetears and dekoyl!!!

still dont understand the 'k' part of dekoyl's.
lol

q. what is the coefficient of x^3 in same question above? (what is the terms)
Title: Re: binomial no calc
Post by: TrueTears on March 19, 2009, 09:06:04 pm: Exactly the same principle as how I did $x^2$ , try it and if you don't get it, I'll write up my working :)
Title: Re: binomial no calc
Post by: lacoste on March 19, 2009, 09:07:22 pm: no wait i get it now, thanks man!! subing the 'x^2 in' cheers, we do that for all the tricky q's? yeh where theres an x at the bottom
Title: Re: binomial no calc
Post by: TrueTears on March 19, 2009, 09:08:33 pm: Quote from: lacoste on March 19, 2009, 09:07:22 pm
no wait i get it now, thanks man!! subing the 'x^2 in' cheers, we do that for all the tricky q's? yeh where theres an x at the bottom
yeah but if you have basic equations which you can take a 'x' out and have a linear factor inside, then don't bother with substitution. But that question, no matter how much you manipulated it, you can never get a linear expression. So just use substitution :P
Title: Re: binomial no calc
Post by: lacoste on March 19, 2009, 09:12:14 pm: It didnt work, i tried 3rd term and fifth row, switched both of the terms and used a calc to check and got 720/x
and 1080x
Title: Re: binomial no calc
Post by: lacoste on March 19, 2009, 09:20:00 pm: how do i do the quick indices check, i havent fully understood it?

can someone please explain?
Title: Re: binomial no calc
Post by: TrueTears on March 19, 2009, 09:26:30 pm: Quote from: lacoste on March 19, 2009, 09:12:14 pm
It didnt work, i tried 3rd term and fifth row, switched both of the terms and used a calc to check and got 720/x
and 1080x

oh damn, I'm beginning to have doubts about my method now lol

Frankly I am stomped atm, I don't see how you can work out the coefficient of $x^3$ without expanding... hmm maybe dekoyl or mao can help :)
Title: Re: binomial no calc
Post by: dekoyl on March 19, 2009, 09:29:17 pm: Okay um I'll have a go.

Possible $x^{3}$ combo is only $\frac{kx^4}{x}$ .

So looking at Pascal's triangle.

$(5)\left(\frac{2}{x}\right)^1\left(3x})^{4}$
$=\frac{810x^4}{x}$
$=810x^{3}$

So $x^3$ coefficient is 810.

Is this right? I'm not sure how to do it without expanding o_O

Edit: Just checked on calc after numerous syntax errors =S. Should be right
Title: Re: binomial no calc
Post by: TrueTears on March 19, 2009, 09:32:39 pm: I suggest dekoyl's method, works perfectly. :)
Title: Re: binomial no calc
Post by: lacoste on March 19, 2009, 09:38:22 pm: thats right dekoyl but how does (kx^4)/x^3 relate to pascals triangle, where do you look in the triangle?
Title: Re: binomial no calc
Post by: dekoyl on March 19, 2009, 09:41:55 pm: Quote from: lacoste on March 19, 2009, 09:38:22 pm
thats right dekoyl but how does (kx^4)/x^3 relate to pascals triangle, where do you look in the triangle?
On the previous page I bolded the row I looked at (1,5,...5,1)
$k$ is just a constant. I say $\frac{kx^4}{x}$ because for this question you posted, there will not be a coefficient in the denominator.
I really suck at explaining =P Maybe someone more articulate can explain.
Title: Re: binomial no calc
Post by: lacoste on March 19, 2009, 09:55:51 pm: $x^{3}$ combo is only $\frac{kx^4}{x}$ . <<< I understand this now, your just making anything equal to x^3

"So looking at Pascal's triangle."

What do you mean by look at pascals triange regarding the above of $\frac{kx^4}{x}$ .

where do u go to pascals to get nCr(5,1)?