# Binomial expansions

1. Aug 17, 2010

### thereddevils

1. The problem statement, all variables and given/known data

Find, in the simplest form, the coefficient of x^n in the binomial expansion of (1-x)^(-6).

2. Relevant equations

3. The attempt at a solution

i am not sure how to go about with this.

2. Aug 17, 2010

### Willian93

are u sure you have to find a cofficient contain x^n?

because u should have a specific value for n, so that you could find the cofficient infront of it, or you should at least have which term you are looking for.

3. Aug 18, 2010

### thereddevils

Yes, that's the question. Maybe it's asking for the coefficient of for any term in the expression.

4. Aug 18, 2010

### HallsofIvy

Staff Emeritus
Since the exponent, -6, is not a positive integer, you will need to use the generalized binomial series:
$$(a+ b)^m= \sum_{k=0}^\infty \frac{m(m-1)\cdot\cdot\cdot(m-k+1)}{k!}a^kb^{m-k}$$

Here, of course, a= 1, b= -x, and m= -6 so this is

$$(1- x)^{-6}= \sum_{k=0}^\infty \frac{-6(-7)\cdot\cdot\cdot(-5-k)}{k!}(-1)^kx^{-6-k}$$

You, apparently, are asked for the coefficient when -6-k= n or when k= -6-n.

5. Aug 18, 2010

### thereddevils

thanks

Is this answer the most simplified?

$$\frac{-6(-7)...(-11-n)}{(-6-n)!}$$

the general formula for binomial series for (a+b)^n is different when n is a positive integer and when n is a fractional or negative value?

$$(a+ b)^m= \sum_{k=0}^\infty \frac{m(m-1)\cdot\cdot\cdot(m-k+1)}{k!}a^kb^{m-k}$$

Does it matter if the powers(k and n-k) for a and b is swapped since a+b is commutative?

This is the continuation of the question:

Hence, find the coefficient of x^6 and x^7 in (1+2x+3x^2+4x^3+5x^4+6x^5+7x^6)^3

Last edited: Aug 18, 2010
6. Aug 19, 2010

### thereddevils

any further hints on this question?