Review question

# What's the sum of the $r^{th}$ bracket in $(1), (2,3), (4,5,6),\cdots$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7240

## Solution

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 $n\dfrac{1}{2} \text{(first term + last term)} = \dfrac{r}{2}\left(\dfrac{r^2-r+2}{2} + \dfrac{r(r+1)}{2}\right) = \dfrac{r(r^2+1)}{2}.$