1. ## Binomial expansions

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

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

Attempt:

The generalised binomial theorem,

$\displaystyle \sum^{\infty}_{k=0}\frac{-6(-6-1)(-6-2)...(-6-r+1)}{r!}(1)^{-6-r}(-x)^r$

So when r=n, the coefficient of x^n will be $\displaystyle -\frac{-6(-6-1)(-6-2)...(-6-r+1)}{r!}$

Can that be further simplified?

And how can i go about in the second part?

2. Originally Posted by hooke
Find, in the simplest form, the coefficient of x^n in the binomial expansion of (1-x)^(-6).

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

Attempt:

The generalised binomial theorem,

$\displaystyle \sum^{\infty}_{k=0}\frac{-6(-6-1)(-6-2)...(-6-r+1)}{r!}(1)^{-6-r}(-x)^r$

So when r=n, the coefficient of x^n will be $\displaystyle -\frac{-6(-6-1)(-6-2)...(-6-r+1)}{r!}$

Can that be further simplified?
The terms all have positive signs, since they contain an even number of negatives. So you can leave out all the negative signs. The first few terms are

$\displaystyle (1-x)^{-6} = 1 + 6x + \frac{(6)(7)}{2!}x^2 + \frac{(6)(7)(8)}{3!}x^3 + \ldots .$

Originally Posted by hooke
And how can i go about in the second part?
Hint: The generalised binomial expansion of $\displaystyle (1-x)^{-2}$ is

$\displaystyle (1-x)^{-2} = 1 + 2x + 3x^2 + 4x^3 + \ldots .$ It's an infinite series, but its first seven terms are the same as those in the expression that you want the cube of.

3. Originally Posted by Opalg
The terms all have positive signs, since they contain an even number of negatives. So you can leave out all the negative signs. The first few terms are

$\displaystyle (1-x)^{-6} = 1 + 6x + \frac{(6)(7)}{2!}x^2 + \frac{(6)(7)(8)}{3!}x^3 + \ldots .$

Hint: The generalised binomial expansion of $\displaystyle (1-x)^{-2}$ is

$\displaystyle (1-x)^{-2} = 1 + 2x + 3x^2 + 4x^3 + \ldots .$ It's an infinite series, but its first seven terms are the same as those in the expression that you want the cube of.
Thanks a lot Opalg, so my answer for the first part is correct except for the extra negative sign?

4. (1-x)^(-6)=1+6x+21x^2+56x^3+...

I am trying to find the general formula for the coefficient of x^(k-1)

Firstly, i will have to start by finding the general formula for 1,6,21,56,...

But how?

5. Originally Posted by hooke
(1-x)^(-6)=1+6x+21x^2+56x^3+...

I am trying to find the general formula for the coefficient of x^(k-1)

Firstly, i will have to start by finding the general formula for 1,6,21,56,...

But how?
$\displaystyle (1-x)^{-6}=\binom{-6}{0}(-x)^0+\binom{-6}{1}(-x)^1+\binom{-6}{2}(-x)^2+.....+\binom{-6}{k-1}(-x)^{k-1}+...$

$\displaystyle =\binom{-6}{0}(-1)^0x^0+\binom{-6}{1}(-1)^1x^1+\binom{-6}{2}(-1)^2x^2+....+\binom{-6}{k-1}(-1)^{k-1}x^{k-1}+...$

6. Originally Posted by hooke
(1-x)^(-6)=1+6x+21x^2+56x^3+...

I am trying to find the general formula for the coefficient of x^(k-1)

Firstly, i will have to start by finding the general formula for 1,6,21,56,...

But how?
The formula $\displaystyle -\frac{-6(-6-1)(-6-2)\cdots(-6-n+1)}{n!}(-x)^n$ for the term in x^n (given in the original post) is correct. There is an even number of minus signs, so they all cancel out, leaving $\displaystyle \frac{6(6+1)(6+2)\cdots(6+n-1)}{n!} = \frac{6\cdot 7\cdot 8\cdots(n+5)}{n!}$ as the coefficient of x^n.

To simplify this further, notice that for n>5 there will be some cancellation between the numbers on the top and bottom of the fraction. For example, the coefficient of x^8 is $\displaystyle \frac{\rlap{/}6\cdot \rlap{/}7\cdot \rlap{/}8\cdot 9\cdot 10\cdot 11\cdot 12\cdot 13}{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot \rlap{/}6\cdot \rlap{/}7\cdot \rlap{/}8}$. In fact, for a general n>5, all that survives will be the last five numbers in the numerator and the first five numbers in the denominator. So the coefficient of x^n is $\displaystyle \boxed{\tfrac1{120}(n+1)(n+2)(n+3)(n+4)(n+5)}$ (and you can check that that formula also woks for $\displaystyle n\leqslant5$).