1. ## 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)

2. 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 :-)

3. 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?

4. 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.

5. 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)

6. 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.

7. 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 :

$(1+x)^n=\sum_{k=0}^{\infty} C_n^k x^k$

8. 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

9. 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.

10. 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!)].

11. You can also do it the good ol' way

...

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

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

@Plato: is that the multinomial theorem?

12. 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!)].
I got 192 as the answer, is this correct?

13. 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!]

14. 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?

15. 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

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

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