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First I have to expand it then find the fourth term, right? But how do I do that?

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2  
Binomial theorem. –  Henning Makholm Apr 10 '12 at 22:55
    
I understand that. But this is the first time I'm doing something like this. There's also a notation with the big C. I'm not familiar with it. –  David Apr 10 '12 at 22:57
    
I recommend reading en.wikipedia.org/wiki/Binomial_theorem –  Belgi Apr 10 '12 at 22:58
    
There is also Pascal's Triangle. –  Peter Grill Apr 10 '12 at 23:13
3  
Define fourth term. –  Aryabhata Apr 11 '12 at 1:17

3 Answers 3

up vote 5 down vote accepted

If you're aware of the binomial theorem, you have that

$$(a-4b)^9=\sum\limits_{k=0}^9 {9 \choose k}a^k (-1)^{9-k} (4b)^{9-k}$$

$$(a-4b)^9=\sum\limits_{k=0}^9 {9 \choose k}a^k (-1)^{k+1} (4b)^{9-k}$$

The fourth term is then the case $k=3$, from which

$$c_3 = {9 \choose 3}a^3 (-1)^{4} (4b)^{6}=344064a^3b^{6}$$

Since you might probably want the other symetrical coefficient, it is

$${9 \choose 6}a^6 (-1)^{7} (4b)^{3}=-5376a^6 b^{3}$$

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Minor correction: I think you accidentally wrote $a^3b^6$ instead of $a^6b^3$. –  E.O. Apr 10 '12 at 23:44
    
@EmileOkada I looked at it from the other side. –  Pedro Tamaroff Apr 11 '12 at 0:24

The binomial expansion in the general form is $$(x+y)^n={n\choose0}x^n+{n\choose1}x^{n-1}y+\cdots+{n\choose n-1}xy^{n-1}+{n\choose n}y^n=\sum_{k=0}^{n}{n\choose k}x^{n-k}y^k$$

In your case $n=9$, $x=a$ and $y=-4b$. Since the fourth term is when $k=3$ you get $${9\choose3}a^6(-4b)^3=84a^6(-64)b^3=-5376a^6b^3$$

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The Binomial Theorem tells us that $$(x+y)^n=\sum_{i=0}^n {n\choose i}x^iy^{n-i}.$$

But what the heck is that? Ok, lets start with $(x+y)^n = \underbrace{(x+y)(x+y)\cdots(x+y)}_{\text{n times}}$. Now, all we have to do is expand this out! So basically each term in our sum is going to be the result of picking an $x$ or a $y$ from each element in this product. For example, if we choose the $x$ from every term we'll end up with $x^n$. In fact, there are no other ways to get $x^n$ because we need all of the $x$s, there's no choice involved. What if we choose $n-1$ $x$s and $1$ $y$? Then the $y$ could come from any of the terms in the product, so there are $n$ ways to choose the $y$, and then we have no choice left for the $x$s. This means that in the expanded formula we'll have $\underbrace{x^{n-1}y+x^{n-1}y+\cdots+x^{n-1}y}_{\text{n times}} = nx^{n-1}y$. Now we see that for each term, the coefficient is simply the number of different ways to choose that number of $x$s and $y$s from the product. This is exactly what $n \choose i$ means, and we compute it by $${n\choose i} = \frac{n!}{i!(n-i)!}.$$

That was a long explanation, but basically what it boils down to is the fourth term of $(a−4b)^9$ is, from the binomial theorem, $${n\choose i}x^{n-i}y^i = {9\choose3}(a)^6(-4b)^3.$$

(Here $i$ is $3$ because the first term has $i =0$).

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I think you could make the shorter question. In how many ways can we write $x^{n-k}y^{k}$ for $k=0,1,\dots,n$? –  Pedro Tamaroff Apr 20 '12 at 0:34

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