Coefficient of x^r in Expansion of (1+x)(1-x)^n

[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 x^r 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.

 

[Simon Bridge]

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?

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?

 

[micromass]

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.

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 $m>n$.

 

[AlephZero]

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?

Yes.

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?

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.

 

[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?

 

[Appleton]

Thank you so much for clarifying that for me.

 

[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...

 

[Simon Bridge]

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?

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