Given the binomial expansion of $(1+x)^n$, can we find $x$ and $n$?

Given the binomial expansion of $(1+x)^n$, can we find $x$ and $n$?

Question

The binomial expansion of $(1+x)^n$ is

$$1-\dfrac{1}{2}\times \dfrac{1}{3} + \dfrac{\frac{1}{2}\times\frac{3}{2}}{1\times 2}\left(\dfrac{1}{3}\right)^2- \dfrac{\frac{1}{2}\times\frac{3}{2}\times\frac{5}{2}}{1\times 2\times 3}\left(\dfrac{1}{3}\right)^3+...$$

Determine the values of $x$ and $n$.

Suggestion

The binomial expansion of $(1+x)^n$ is

$$1-\dfrac{2}{3}\times \dfrac{1}{3} + \dfrac{\frac{1}{2}\times\frac{3}{2}}{1\times 2}\left(\dfrac{1}{3}\right)^2- \dfrac{\frac{1}{2}\times\frac{3}{2}\times\frac{5}{2}}{1\times 2\times 3}\left(\dfrac{1}{3}\right)^3+...$$

Determine the values of $x$ and $n$.

Could we write out the expansion of $(1+x)^n$ and compare the terms with what we are given?

Solution

We can write down the binomial expansion of $(1+x)^n$ as

$$1+\dfrac{n}{1!}x + \dfrac{n(n-1)}{2!}x^2+ \dfrac{n(n-1)(n-2)}{3!}x^3+...$$

This is true for all real values of $n$, although there are conditions on $x$.

If $n$ is a positive integer, the expansion terminates, while if $n$ is negative or not an integer (or both), we have an infinite series that is valid if and only if $\big \vert x \big \vert < 1$.

We can now compare this with the series we are given. The increasing powers of $\dfrac{1}{3}$ strongly suggest that $x = \dfrac{1}{3}$.

What can we now say about $n$? If we look at the $x^2$ term, we need $\dfrac{1}{2}\times\dfrac{3}{2} = n(n-1)$. The only ways this can happen are if $n = -\dfrac{1}{2}$ or $n = \dfrac{3}{2}$.

The $x^3$ term tells us $-\dfrac{1}{2}\times\dfrac{3}{2}\times\dfrac{5}{2} = n(n-1)(n-2)$, which must mean $n = -\dfrac{1}{2}$.

We can easily check that this provides the other terms of the series successfully.

Thus $x = \dfrac{1}{3}, n = -\dfrac{1}{2}$.