First I’m going to assume you are competent with binomial expansions of the form (p+q)n n∈N
that is,
(p+q)ⁿ = nC0 pⁿ + nC1 pⁿ⁻¹q + nC2 pⁿ⁻²q²+ nC3 pⁿ⁻³q³ + .... + nCn-r pʳqⁿ⁻ʳ +.. + nCn-1 pqⁿ⁻¹ + qⁿ
and nCr = n! / r! (n-r)!
Expanding the notation,
(1+x)ⁿ = pn +n pⁿ⁻¹ q +n(n-1) pⁿ⁻² q²/2! + n(n-1)(n-2) pⁿ⁻³3q³ /3! + ..+ n(n-1)p²qⁿ⁻²/2! + npqⁿ⁻¹ +qⁿ
I shall assume you can see the symmetry.
Further, you should see that each new coefficient can be calculated from the last one:
nCr = nCr-1 (n-r)/r
In statistics, probability calculations are based upon expansions of (p+q)ⁿ, where p+q = 1. Many students get stuck here, seeing that 1ⁿ is obviously one and therefore, they think, of no interest. Au contraire, mon brave; adding up to one simply confirms that the probabilities are all there and it is the way the ‘one’ is divided up that tells you the distribution of the various probabilities. That’s the interesting bit, the distribution.
I put examples in the Lower School Binomial. The usage is in Hypothesis testing. There I have assumed you recognise “X is a variate that behaves as a binomial distribution with n items of probability p” as being written X~B(n,p) and that the mean is np (that’s the expectation) and the variance is npq. I have not proved that on these pages. Yet.
