Binomial expansion for any n

Question

I teach A-Level maths, and in the second year we do the general binomial expansion, which is even provided for the students in the formula book.

For values of $n$ that are not positive integers: (I realise this form does also work for postive integers as well.)

$$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2+\ldots$$

Every textbook I have read says that if you then need to expand $(a+x)^n$, then you have to do $a^n(1+\frac{x}{a})^n$, expand the bracket and then multiply the constant back in. You have to have it in the form $(1+\mathrm{f}(x))^n$. You can then give the validity of the expansion $\left|\frac{x}{a}\right|<1 \Rightarrow |x| < |a|$.

We discuss the convergence issue, that in order for the infinite sequence to converge we need the '$x^n$' terms to converge to zero, and hence the validity statement.

So is there any reason why we cannot miss out that step and just do the following, which will result in exactly the same expansion?

$$(a+x)^n = a^n + na^{n-1}x + \frac{n(n-1)}{2!}a^{n-2}x^2+\ldots+\frac{n(n-1)\ldots(n-r+1)}{r!}a^{n-r}x^r+\ldots$$

This combines the second year learning, with the version they are given in the first year for positive integer values only:

$$(a+b)^n = a^n + ^nC_1a^{n-1}b + ^nC_2a^{n-2}b^2 + \ldots + ^nC_ra^{n-r}b^r +\ldots + b^n$$

Is it always provided using the first method just in order to simplify the learning for students? While I think I'll keep teaching it the way it is presented in their textbooks, I don't want to tell students they can't do it like that if they're already using that approach.

Answer 1

The reason why you need to do it the first way is because combinatorics are only defined within the domain of the natural numbers. Therefore, the notation you are using for the "simplified" approach is garbleglorp and will confuse your students. The first approach is the "generalized binomial theorem" and cannot be proven using combinatorics. You, therefore, need to prove it by showing the series is convergent for negative and fractional rational exponents as well. This is proven to be true for all rational numbers n iff the equation is (1 + x)ⁿ and |x| is small.

EDIT: I misread your question.

Answer 2

As a pedagogical point, the series expansion

$$ (1+x)^n = 1 + x + \frac{n(n-1)}{2} x^2 + \ldots $$

is somewhat simpler. This has some pedagogical value:

• When doing a "complicated" calculation, it's best to make the things you're calculating with as simple as possible
• If you're going to memorize something, memorizing the simpler one above is arguably easier than trying to memorize $(a+x)^n$
• It's a nice example in the art of reducing the general case to a special particular case
• The actual calculus work is purely single-variable calculus, rather than multi-variable calculus with all but one variable held constant

As an aside, you may like to take this opportunity to introduce generalized binomial coefficients $\binom{n}{r}$ in the case where $n$ is no longer restricted to be a nonnegative integer.