Dear Uncle Colin,

I’m supposed to find the coefficient of $x^0$ in the expansion of $\left(x^2 + \frac{1}{x}\right)^{15}$ – I couldn’t figure it out. Any advice?

- Combinations Of Numbers Seem Tough And Nasty, Though

Hi, CONSTANT, and thanks for your message! As per, I can see several approaches.

Approach 1: a sledgehammer

A perfectly sensible approach here is to work out the first few terms: $\nCr{15}{0} x^{30} + \nCr{15}{1}x^{27} + \nCr{15}{2}x^{24} + \dots$ – oh! It looks like the powers are going down by three each time.

To get the zeroth power, I’ll have $\nCr{15}{10}x^0$, and $\nCr{15}{10} = 3,003$.

Approach 2: pulling out a factor

$\left(x^2 + \frac{1}{x}\right)^{15} = \left( \frac{1}{x^{15}}\right)\left(x^3 + 1\right)^{15}$.

To get a constant term when the $\frac{1}{x^{15}}$ is multiplied back in, we need to find the $x^15$ term in the second factor – again, that comes from the 10th term, and the coefficient is again $\nCr{15}{10} = 3,003$.

Approach 3: apply the formula

Knowing that the $k$th term of $(a+b)^n$ is $\nCr{n}{k}a^{n-k} b^{k}$ can work nicely here.

We have $a = x^2$, $b = x^{-1}$ and $n=15$, so the $k$th term is $\nCr{15}{k}x^{30-2k} x^{-k}$, or $\nCr{15}{k} x^{30-3k}$.

The power is zero when $k=10$, giving - yet again - a coefficient of $\nCr{15}{10} = 3,003$.

Interesting number, 3,003. It shows up in Pascal’s Triangle eight times. No-one knows whether a number (other than 1) appears infinitely often.

Hope that helps!

- Uncle Colin