Canonical derivation of the mean and variance of the binomial distribution

The mean and variance of the binomial distribution can be derived most easily using a simple indicator function approach but, in the course of studying a stochastic process involving the binomial distribution, I became interested in deriving the mean and variance from the canonical probability-weighted sum of random variable values. I found it instructive to work out how to do this and want to record the relevant manipulations here.

For the binomial distribution $B(N, p)$ with $N$ a positive integer and $0 < p < 1$ , the underlying random variable $X$ is one that counts the number of successes in the first $N$ trials of a Bernoulli sequence with probability $p$ of success on any trial and a probability $q = 1 - p$ of failure. The probability function for $n$ successes in $N$ trials is

$P(X = n) = C(N, n) p^n q^{N-n} \quad \quad \quad \quad (1)$

where

$C(N, n) = \frac{N!}{n! (N-n)!} \quad \quad \quad \quad (2)$

The mean and variance of the random variable $X$ can be obtained almost trivially by expressing $X$ as a sum of indicator functions $I_k$ for $k = 0, 1, \ldots, N$ , where each indicator takes the value 1 meaning success with probability $p$ and the value 0 meaning failure with probability $q$ :

$X = \sum_{k=1}^{N} I_k \quad \quad \quad \quad (3)$

Given that

$E[I_k] = p \quad \quad \quad \quad (4)$

and

$V[I_k] = E[I_k^2] - E^2[I_k] = pq \quad \quad \quad \quad (5)$

it follows immediately that

$E[X] = Np \quad \quad \quad \quad (6)$

and

$V[X] = Npq \quad \quad \quad \quad (7)$

However, it is also possible to obtain these expressions by manipulating the usual defining sums for $E[X]$ and $E[X^2]$ using the probability functions in (1) above, i.e.,