The positive integers are bracketed as follows: \[(1), (2,3), (4,5,6),\dotsc ,\] where there are \(r\) integers in the \(r^{th}\) bracket.

Find expressions for the first and last integers in the \(r^{th}\) bracket.

Let’s look at the sequence \(u_1, u_2, u_3, \cdots\) formed by the first term in each bracket. This is \[ 1,2,4,7,11,\cdots\] where the \(r^{th}\) term \(u_r\) is given by \(1+1+2+3+\cdots+(r-1)\).

*You may assume that the sum of the first \(n\) positive integers is \(\frac{1}{2} n (n+1)\).*

Using the formula for \(1 + 2 + 3 + \cdots\) given in the questions, we can write this as \[1+ \dfrac{r(r-1)}{2} = \dfrac{r^2-r+2}{2}.\]

*Check: when \(r = 4, \dfrac{r^2-r+2}{2} = 7.\)*

Thus \(u_r = \dfrac{r^2-r+2}{2}\).

The last term in the \(r^{th}\) bracket must be \[\dfrac{r^2-r+2}{2} + r - 1 = \dfrac{r(r+1)}{2}.\]

*Check: when \(r = 3, \dfrac{9+3}{2} = 6\).*

Find the sum of all the integers in the first 20 brackets.

We know the last integer of the \(20^{th}\) bracket is \(\dfrac{20(21)}{2} = 210\).

So to find the sum of the first 20 brackets, we just need to add all positive integers up to \(210\).

This gives us \[\frac{1}{2} {210} (210+1) = 22\, 155.\]

Prove that the sum of the integers in the \(r^{th}\) bracket is \(\dfrac{1}{2}r(r^2+1)\).

The \(r^{th}\) bracket starts with \(\dfrac{r^2-r+2}{2}\) and ends with \(\dfrac{r(r+1)}{2}\).

This is an arithmetic series with \(a = \dfrac{r^2-r+2}{2}, n = r\) and \(d = 1\).

Thus the sum of the terms is \[n\dfrac{1}{2} \text{(first term + last term)} = \dfrac{r}{2}\left(\dfrac{r^2-r+2}{2} + \dfrac{r(r+1)}{2}\right) = \dfrac{r(r^2+1)}{2}.\]