Binomial expansion

by Appleton
Tags: binomial, expansion
 P: 40 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.
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 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?
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 Quote by Appleton 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##.

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Binomial expansion

 Quote by Appleton 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.
 Homework Sci Advisor HW Helper Thanks ∞ P: 12,356 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?
 P: 40 Thank you so much for clarifying that for me.
 P: 5 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...
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