This is a question about a simple random walk problem where we want to measure the average distance from the start in various experiments of $N$ steps each. If $d$ is distance moved during one such experiment, then: $\langle d\rangle = \langle(a_1 + a_2 + a_3 + \cdots + a_N)\rangle = \langle a_1\rangle + \langle a_2\rangle + \langle a_3\rangle + \cdots + \langle a_N\rangle$ $ = 0$
This clearly seems to be the answer the problem is lookng for. Why then do we try to find the average of the square of $d$ as:
$$\langle d^2\rangle = \langle (a_1 + a_2 + a_3 + \cdots + a_N)^2\rangle = \langle(a_1 + a_2 + a_3 + \cdots + a_N) (a_1 + a_2 + a_3 + \cdots + a_N)\rangle = N \text{ ?}$$
Why are we squaring the distance for no obvious reason? I have read multiple sources that say the square is taken to get rid of the minus sign. But that does not make sense to me because the average of two negative numbers, say, $-1$ and $1$ is meant to be $0.$
To begin with, consider the difference between distance and displacement. The displacement is a vector, the distance is a nonnegative number. The expected displacement of a symmetric random walk is always $0$, and thus is not an interesting quantity to look at. The expected distance is typically positive, and grows with time; we may be interested in how fast it grows.
The displacement after $n$ steps is $a_1+\dots +a_n $. The distance from $0$ after $n$ steps is the magnitude of displacement: $|a_1+\dots +a_n |$. Unfortunately, expressing the expected value of $|a_1+\dots +a_n |$ in terms of the distribution of $a_i$ is quite difficult. It is much easier to deal with the square, because cross-products contribute $0$ by virtue of the steps being independent.