# Slick Summing

Watch the video to see how Charlie works out the sum. Can you adapt his method?

## Problem

*Slick Summing printable worksheet*

In the video below, Charlie works out $1+2+3+4+5+6+7+8+9+10$.

Can you see how his method works?

How could you adapt his method to work out the following sums?

$1 + 2 + 3 + \dots + 19 + 20$

$1 + 2 + 3 + \dots + 99 + 100$

$40 + 41 + 42 + \dots + 99 + 100$

Can Charlie's method be adapted to sum sequences that don't go up in ones?

$1 + 3 + 5 + \dots + 17 + 19$

$2 + 4 + 6 + \dots + 18 + 20$

$42 + 44 + 46 + \dots + 98 + 100$

Can you find an expression for the following sum?

$1 + 2 + 3 + \dots + (n - 1) + n$

**Notes and Background**

If you enjoyed this problem you may be interested to read the article Clever Carl, which tells the story of the young Gauss working on sums like the ones in this problem. You may also be interested in this video which shows a dozen proofs of the formula of the sum of the first n positive integers.

## Getting Started

Take a look at Picturing Triangular Numbers first.

## Student Solutions

We've received lots of good solutions to this problem. Thanks to Lasse (from St. James Senior Boys' School), Fred (from Wembley High Technology College), Nishil (from Cambridge School), Rishi (from St. Matthew's Primary School), Zachary (from Sonoran Science Academy), Kaushal (from Overseas Family School), Rukkshana and Annie (from Fern Avenue Public School), Julia and Grace (from Inter-Lakes), Julian (from British School Manila), Poppy (from Queen Elizabeth's Grammar School in Horncastle), Ci Hui (from Queensland Academy for Science Mathematics and Technology, Australia), Shaunak (from India) and Qiun (from China) for all the great answers.

Poppy used a diagram to represent Charlie's method as part of working on this problem:

Here's Zachary's solution to the first thee sums in the problem:

First, let's work out a solution to 1+2+3+...+10.

When we pair the first and last term, 2nd and 9th term, 3rd and 8th term, ..., we get sums of 11. We can list the pairs:

- 1 and 10
- 2 and 9
- 3 and 8
- 4 and 7
- 5 and 6

There are 5 pairs of sums that add up to 11, and so we get a sum of 5*11 = 55.

Now, we need to find 1+2+3+...+20. Now listing out the terms would be a bad idea, so let's think about how many terms that was in the sum of 1-10. There were 5 pairs in 1-10. So, if the same rules apply, then there would be 10 pairs in the sum of 1-20. Now, we find the sum by adding the first and the last terms of the sequence. 1+20= 21. We have 10 pairs of 21, and the sum is 10*21= 210.

We will do the same thing for 1-100. 100/2= 50 pairs. 1+100= 101. 50 pairs of 101 is 50*101= 5050.

Now, we have the sum of 40-100. This is harder to solve, because the first term isn't 1. So, let's start by finding the number of terms. First, we must realize that in the set of numbers from n to m, there are m-n+1 terms. So here we have the numbers from 40 to 100, so there are 61 terms. We use the same method, even though there are not an even number of terms. 61/2=30.5. 100+40= 140. 140*30.5= 4270.

Brilliant!

Shaunak offered an alternative to this method for the sum fo 40-100. Their calculation was

$5050-39 \times \dfrac{40}{2}$

How would you explain why this will give the same answer as Zachary's method?

Lasse, Ci Hui and Shaunak pointed out that sums like this are called 'arithmetic series' - well spotted. Kaushal proved a general formula for us:

The sum 1 + 2 + 3 + 4 ... + n-1 + n can be rearranged:

(1 + n) + (2 + (n - 1)) + (3 + (n - 2)) +...

where each term is in the form (x + (n - (x-1)). This simplifies to:

(1 + n) + (1 + n) + (1 + n) +...

The number of ''(1 + n)''s that are added is n/2 since (x + n - (x-1)) is always n + 1. Therefore,

$1 + 2 + 3 + 4 +\dots + n-1 + n = \dfrac{n^2 + n}{2}$.

Julian gave us the following answers:

- 1+3+5+...+17+19 is (20*10)/2 which is 100
- 2+4+6+...+18+20 is (22*10)/2 which is 110
- 42+44+46...+98+100 is (142*30)/2 which is 2130

Great - but we then asked where these come from.

Aarna, from Wallington High School for Girls, UK, gave a solution with further details for the sum $42 + 44+ \cdots + 98+100.$

First she found the sums of pairs of numbers:

$42+100=142$

$44+98=142$

$46+96=142$

$\vdots$

$98+44=142$

$100+42=142$

Aarna then worked out the number of these pairs by counting the number of even numbers from 42 to 100. She found that there were five even numbers in each group of ten numbers such as $42-50$ and there were six of these groups from $42-100$. As she explains,

The total number of even numbers $42 - 100 =6\times 5=30$.

$30=$ number of pairs

$30 \times 142=4260$

$4260 \div 2 = 2130$

So: $42+44+46+\cdots +98+100=2130.$

Other approaches involved more algebra. Ci Hui showed how the answers James had found came from the formula for the sum of an arithmetic series. If the first term is $a$ and the common difference is $d$, then the sum of the first $n$ terms is $\dfrac{n}{2} (a + a +(n-1)d)$ or $\dfrac{n}{2} (2a +(n-1)d)$. Can you connect this formula to Charlie's method?

Although this formula can be used for all these calculations, Shaunak found specific formulae for the sums of the first $k$ odd numbers and the first $k$ even numbers. As she explains:

$n$ is the last term of the progression. For AP (arithmetic progression) with a difference of 2, containing odd numbers, the formula is $\left(\dfrac{n + 1}{2}\right)^2$:

$1 + 3 + 5 + … + 2k – 1$

$2k – 1 + … + 5 + 3 + 1$

$\dfrac{(2k + 2k + 2k + … + 2k + 2k)}{2} = \dfrac{k(2k)}{2} = \dfrac{2k^2}{2} = k^2= \left(\dfrac{n + 1}{2}\right)^2$.

For AP with a difference of 2, containing even numbers, the formula is $\left(\dfrac{n}{2}\right) \left(\dfrac{n}{2} + 1\right)$:

$2 + 4 + 6 + … + 2k – 2 + 2k$

$2k + 2k – 2 + … + 6 + 4 + 2$

$\dfrac{(2k + 2 + 2k + 2 + 2k + 2 + … + 2k + 2 + 2k + 2)}{2} = \dfrac{k(2k + 2)}{2}$

$= \dfrac{(2k^2 + 2k)}{2} = 2\dfrac{(k^2 + k)}{2} = k^2 + k =\left( \dfrac{n}{2}\right)^2 + \dfrac{n}{2} = \left(\dfrac{n}{2}\right)\left(\dfrac{n}{2} + 1\right)$.

Well done to everyone who sent in solutions for this problem!

## Teachers' Resources

### Why do this problem?

This problem provides an introduction to summing simple arithmetic series. By considering a particular case, students are invited to attend to the structure and see where the general method for summing such series comes from.

The Stage 5 problem Speedy Summations poses more challenging questions using $\sum$ notation.

### Possible approach

*This printable worksheet may be useful: **Slick Summing**.*

You may wish to show the video, in which Charlie works out $1+2+3+4+5+6+7+8+9+10$ in silence, or you may wish to recreate the video for yourself on the board.

"With your partner, discuss what you saw. Can you recreate the video for yourselves? Can you explain what happened?"

Once students have had a chance to make sense of the video:

"I wonder if you could adapt Charlie's method to work out some similar questions?

Try $1 + 2 + 3 + \dots + 19 + 20$ first."

While students are working, write up a few more questions on the board, such as

$1 + 2 + 3 + \dots + 99 + 100$

and

$40 + 41 + 42 + \dots + 99 + 100$

*Circulate to check for misconceptions - one common one is that the sum from 1 to 20 will be twice the sum from 1 to 10, and for the last example above, some students may assume the sequence has 60 terms.*

Bring the class together to discuss what they have done.

"I wonder whether Charlie's method could be adapted for sequences that don't go up in ones?"

$1 + 3 + 5 + \dots + 17 + 19$

$2 + 4 + 6 + \dots + 18 + 20$

$42 + 44 + 46 + \dots + 98 + 100$

"In a while I'm going to give you another question like these and you'll need to have developed a strategy to work it out efficiently."

While students are working, listen out for useful comments that they make about how to work out such sums generally. Then bring the class together to share answers and methods for the questions they have worked on.

Make up a few questions like those above, and invite students out to the board to work them out 'on the spot', explaining what they do as they go along. This might be an opportunity to invite more than one student to the board at the same time to work on the same question.

Next, invite students to create a formula from their general thinking:

"Imagine summing the sequence $1 + 2 + \dots + (n-1) + n$.

Can you use what you did with the numerical examples to create an expression for this sum?"

Give students time to think and discuss in pairs and then build up the formula together.

### Key questions

What can you say about the sum of the first and last, and the second and penultimate terms of these sequences?

How do you know these sums of pairs will always be the same?

### Possible support

Picturing Triangle Numbers offers a pictorial representation of the sum of the first $n$ integers

### Possible extension

Speedy Summations contains some challenging follow-up questions.

### .