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 \(n\).
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 \(x = \dfrac{1}{3}, n = -\dfrac{1}{2}\).
