Finding the $p/q$ th term in a series 
Introduction:

The rational numbers can be expressed in this simple sequence$$\tfrac 11,\space\tfrac 21,\space\tfrac 12,\space\tfrac 31,\space\tfrac 22,\space\tfrac 13,\space\tfrac 41,\space\tfrac 32,\space\tfrac 23,\space\tfrac 14,\space\ldots\ldots$$
And it can be shown that $p/q$ is the $\left[\tfrac 12(p+q-1)(p+q-2)+q\right]$th term of the series.

Questions:



*

*How can you prove that $p/q$ is the $\left[\tfrac 12(p+q-1)(p+q-2)+q\right]$th term?

*Can you justify and explain each step?



Observations:

I've noticed that the sum $p+q$ has $n-1$ fractions for $p+q=n$. $$\begin{align*} & \frac pq=\frac 11\qquad p+q=2\qquad 1\text{ fraction}\\ & \frac pq=\frac 21\qquad p+q=3\\ & \frac pq=\frac 12\qquad p+q=3\qquad2\text{ fractions}\end{align*}$$
And on and on. Is there a way to exploit this?
 A: 
It can be noticed that $\frac{p}{q}$ lies in the $p^{\text{th}}$ row and $q^{\text{th}}$ column. It's discernible that in order to figure out which diagonal to count along to get your required fraction (by seeing which denominator is required), you need to add successive natural numbers to get to the diagonal which had the fraction you need. 

From your observations, to get to the diagonal required to ascertain the position of $\frac{p}{q}$ - i.e. diagonal with fractions in the form of $\frac{a}{b}$ where $a+b=p+q=n$, - you need to first go through $(1+2+3\cdot\cdot\cdot+(p+q-2))=x$ numbers, where $x=T_{(n-2)}=(n-2)^{\text{th}}$ triangular number,
 and then through $q$ numbers to arrive at your fraction.
Therefore, you need to go through $\phi$ numbers, where: $$\phi=q+T_n=q+T_{(p+q-2)}=q+{p+q-1 \choose 2}=q+\frac{(p+q-1)(p+q-2)}{2}$$
Just as required.
A: Yes, there is a way to exploit this. For the first value with $p+q=n$, we will find them at position $1+(1+2+...+(n-2))=1+\frac{(n-2)(n-1)}{2}$ from your observation. From there, we now have to go along until we get $q$ in the denominator, and thus we must go over another $q-1$ steps to find the position of $\frac{p}{q}$: $1+\frac{(n-2)(n-1)}{2}+q-1=\left[\tfrac 12(p+q-1)(p+q-2)+q\right]$
