Binomial expansion

1. Jun 29, 2014

Appleton

I am puzzled by the following example of the application of binomial expansion from Bostock and Chandler's book Pure Mathematics:

If n is a positive integer find the coefficient of xr in the expansion of (1+x)(1-x)n as a series of ascending powers of x.

$(1+x)(1-x)^{n} \equiv (1-x)^{n} + x(1-x)^{n}$

$\equiv\sum^{n}_{r=0} { }^{n}C_{r}(-x)^{r} + x\sum^{n}_{r=0} { }^{n}C_{r}(-x)^{r}$

$\equiv\sum^{n}_{r=0} { }^{n}C_{r}(-1)^{r} x^{r}+ \sum^{n}_{r=0} { }^{n}C_{r}(-1)^{r}x^{r+1}$

$\equiv [1-{ }^{n}C_{1}x+{ }^{n}C_{2}x^{2}...+{ }^{n}C_{r-1}(-1)^{r-1} x^{r-1}+{ }^{n}C_{r}(-1)^{r} x^{r}+...+(-1)^{n}x^{n}]$

$+[x-{ }^{n}C_{1}x^{2}+...+{ }^{n}C_{r-1}(-1)^{r-1} x^{r}+{ }^{n}C_{r}(-1)^{r} x^{r+1}+...+(-1)^{n}x^{n+1}]$

$\equiv\sum^{n}_{r=0} [{ }^{n}C_{r}(-1)^{r} + { }^{n}C_{r-1}(-1)^{r-1}]x^{r}$

The 4th and 5th line seemed a peculiar way of writing it. Were they just trying to demonstrate how the second series is always one power of x ahead?

The last expression seems to require a definition of ${ }^{n}C_{-1}$ which hasn't been defined in the book so I'm guessing I have misunderstood something. Could someone please explain this for me?
Apologies for any typos, I'm using a mobile. Very fiddley.

2. Jun 29, 2014

Simon Bridge

That's what it looks like to me - the author is making a step in the calculation explicit.

Do you see how the last line is derived from the one before it?

Notes:
...everything from the third "equivalence" sign to (but not including) the fourth one is all one line of calculation.
Do Bostock and Chandler number their working, their equations?

3. Jun 29, 2014

micromass

Staff Emeritus
You likely didn't misunderstand anything, the book just has been incomplete. The book should have mentioned that we define ${}^nC_m = 0$ for $m< 0$ and $m>n$.

4. Jun 29, 2014

AlephZero

Yes.

I think the book is a bit careless there. ${}^{n}C_{k}$ is normally only defined for $0 <= k <= n$. But the only "sensible" defintiion when $k < 0$ or $k > n$ is zero. If you define ${}^{n}C_{k}$ as the number of ways to choose objects from a set, there are no ways to choose more than n different objects from a set of n, and you can't choose a negative number of objects. If you define it using Pascal's triangle, any numbers "outside" the triangle need to be 0 to make the formulas work properly.

5. Jun 29, 2014

Simon Bridge

The definition being used should be evident by following the derivation though... looking at the coefficient of x^0, probably why the authors felt they could be a bit sloppy there?

6. Jun 29, 2014

Appleton

Thank you so much for clarifying that for me.

7. Jul 17, 2014

Alicelewis11

The last articulation appears to oblige a meaning of nc−1 which hasn't been characterized in the book so I'm speculating I have misconstrued something. Would someone be able to please clarify this for me?

Expressions of remorse for any typos, I'm utilizing a versatile. Exceptionally fiddle...

8. Jul 18, 2014

Simon Bridge

This question has already been asked and answered - see post #3.