# Binomial Expansion

Printable View

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Jul 23rd 2008, 09:00 AM
Simplicity
Binomial Expansion
The question is to find the first three terms of (1-9x+x^2)^5. It has a hint provided which says: 'Write it as (1-9x(1-x/9))^5 and use the binomial theorem.'

The problem is when using the binomial equation, how should I split it up? What should be my 'x' and what should be my constant?

(Sorry, Latex doesn't work at the moment, so it might appear a bit messy)
• Jul 23rd 2008, 09:33 AM
Chop Suey
You have to first write it as two terms, then you can apply the following:
Code:

n Sigma (n) x^(n-k)y^k k=0  (k)
Btw, by (n)(k) i meant nCk....I miss LaTeX :-(

EDIT: I was kind of vague. n represents the number of terms (like (x+y)^4, n = 4), so if you wanted the third term, you should do this:
Code:

 (n) x^(n-2)y^2 (2)
EDIT2: Example: if i wanted to find the fifth term in the expansion of (x+y)^10, then:
- Identify n = 10
- I need fifth term, so k = 5 - 1 = 4.
- Evaluate :-)
• Jul 23rd 2008, 09:42 AM
Simplicity
Quote:

Originally Posted by Chop Suey
You have to first write it as two terms, then you can apply the following:
Code:

n Sigma (n) x^(n-k)y^k k=0  (k)
Btw, by (n)(k) i meant nCk....I miss LaTeX :-(

EDIT: I was kind of vague. n represents the number of terms (like (x+y)^4, n = 4), so if you wanted the third term, you should do this:
Code:

 (n) x^(n-2)y^2 (2)
EDIT2: Example: if i wanted to find the fifth term in the expansion of (x+y)^10, then:
- Identify n = 10
- I need fifth term, so k = 5 - 1 = 4.
- Evaluate :-)

I know the formula but for:

(1-9x(1-x/9))^5

Should I assume x=1 and y=-9x(1-x/9) and apply the

(x+y)^5 binomial rule?
• Jul 23rd 2008, 09:56 AM
Plato
I have a complaint with the wording of this problem: “Who is to say what are the first three terms in the expansion?” You text material may define it. But I can assure you that there is no general agreement on the answer.
I could argue that the answer is: 1+(-9x)^5+(x^2)^5.
The general term is: [5!/(a!*b!*c!)][(1)^a][(-9x)^b][(x^2)^c] where a+b+c=5.
• Jul 23rd 2008, 10:03 AM
Simplicity
Quote:

Originally Posted by Plato
I have a complaint with the wording of this problem: “Who is to say what are the first three terms in the expansion?” You text material may define it. But I can assure you that there is no general agreement on the answer.
I could argue that the answer is: 1+(-9x)^5+(x^2)^5.
The general term is: [5!/(a!*b!*c!)][(1)^a][(-9x)^b][(x^2)^c] where a+b+c=5.

The first three terms means write upto and including x^2.

As you will get:

[constant]+[coefficient]x+[coefficient]x^2 (The first 3 terms)

(According to textbook)
• Jul 23rd 2008, 10:05 AM
Chop Suey
Quote:

Originally Posted by Air
I know the formula but for:

(1-9x(1-x/9))^5

Should I assume x=1 and y=-9x(1-x/9) and apply the

(x+y)^5 binomial rule?

Yep :-)

(1-9x+x^2)^5 = (1-(9x-x^2))^5

Let u = 9x-x^2

(1-u)^5

Sorry for delay, I dozed off, hehe. :p
• Jul 23rd 2008, 10:06 AM
Moo
Quote:

Originally Posted by Air
I know the formula but for:

(1-9x(1-x/9))^5

Should I assume x=1 and y=-9x(1-x/9) and apply the

(x+y)^5 binomial rule?

Yes :)

Specifically :

$(1+x)^n=\sum_{k=0}^{\infty} C_n^k x^k$
• Jul 23rd 2008, 10:22 AM
Plato
Quote:

Originally Posted by Air
The first three terms means write upto and including x^2.
[constant]+[coefficient]x+[coefficient]x^2 (The first 3 terms)(According to textbook)

Why did you not tell us that to begin with?
1+{(5)(-9x)}+{[5!/(3!2!)](-9x)^2]+5x^2}
1-45x+815x^2
• Jul 23rd 2008, 12:25 PM
Simplicity
I managed to do the question and others similar to the one posted earlier but this is another question that I have come upon and I am struggling on it:

Find the coefficient of x^5 of (1+x+x^2)^4

Can someone explain the method. Thanks in advance. :)
• Jul 23rd 2008, 12:45 PM
Plato
There are only two ways to get x^5 in that expansion:
[(1)^0][(x)^3][(x^2)^1] & [(1)^1][(x)^1][(x^2)^2]
The coefficient of the first is [4!/(0!)(3!)(1!)].
The coefficient of the second is [4!/(1!)(1!)(2!)].
Add those.
• Jul 23rd 2008, 12:58 PM
Chop Suey
You can also do it the good ol' way :p

http://www.gnux.be/latex/data/e12307...968b03924b.png
...

*pants* Must...rest....now...*faints*

P.S. Probability of a mistake in expansion: Very high.

@Plato: is that the multinomial theorem?
• Jul 24th 2008, 06:29 AM
Simplicity
Quote:

Originally Posted by Plato
There are only two ways to get x^5 in that expansion:
[(1)^0][(x)^3][(x^2)^1] & [(1)^1][(x)^1][(x^2)^2]
The coefficient of the first is [4!/(0!)(3!)(1!)].
The coefficient of the second is [4!/(1!)(1!)(2!)].
Add those.

I got 192 as the answer, is this correct?
• Jul 24th 2008, 06:34 AM
Moo
Quote:

Originally Posted by Air
I got 192 as the answer, is this correct?

Hmm ? Did you multiply the two coefficients Plato gave or what ? You're supposed to add them since it's like having ...+*1stCoeff*x^5+..terms of expansion..+*2ndCoeff*x^5+...

and be careful, it's 4!/[0! 3! 1!] and 4!/[1! 1! 2!]
• Jul 24th 2008, 06:40 AM
Simplicity
Quote:

Originally Posted by Moo
Hmm ? Did you multiply the two coefficients Plato gave or what ? You're supposed to add them since it's like having ...+*1stCoeff*x^5+..terms of expansion..+*2ndCoeff*x^5+...

and be careful, it's 4!/[0! 3! 1!] and 4!/[1! 1! 2!]

Oh, I thought it was (4!)/(0!) * 3! * 1! which caused me to get the answer. So, now I get 16 as the answer, is that correct?
• Jul 24th 2008, 06:42 AM
Moo
Quote:

Originally Posted by Air
Oh, I thought it was (4!)/(0!) * 3! * 1! which caused me to get the answer. So, now I get 16 as the answer, is that correct?

Yes (Tongueout)

I'd suggest you to get a software (like maxima or maple, but maple is not free) so that you can check by yourself or even try to check for the other coefficients ^^
Or use a graph calculator... lol
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last