# Speedy summations

Watch the video to see how to add together an arithmetic sequence of numbers efficiently.

## Problem

In the video below, Alison works out \(\sum_{i=1}^{10} i\).

*This video has no sound*

*If you can't watch the video, click below for a description*

Next, Alison writes the numbers from 1 to 10, and then the same set of numbers in decreasing order, 10 to 1, underneath, then adds them in pairs. This gives $10 \times 11=110$.

Finally Alison writes the answer $55$ next to the original sum.

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

- $\sum_{i=1}^{100} i$

- $2+4+6+\dots+96+98+100$

- $\sum_{k=1}^{20} (4k+12)$

- $37+42+47+52+\dots+102+107+112$

- The sum of the first $n$ terms of the sequence $a, (a+d), (a + 2d), (a + 3d) \dots$

After how many terms would $17+21+25+\dots$ be greater than $1000$?

Can you find the sum of all the integers less than $1000$ which are not divisible by $2$ or $3$?

Can you find a set of consecutive positive integers whose sum is 32?

## Getting Started

If you are finding it hard to get started, look at Slick Summing first.

## Student Solutions

Amrit explained how to find a formula for the sum of consecutive numbers, and then adapted it to answer the other questions:

Pair the values as follows:

1 | 2 | 3 | ... | n |

n | n-1 | n-2 | ... | 1 |

Each of the columns sums to $n + 1$, and there are $n$ columns, so the sum

should be $n(n + 1)$. However, as each column is counted twice (once in each

possible order of the two values), the sum is $\frac{n(n + 1)}{2}$.

$2 + 4 + 6 + ... + 96 + 98 + 100 = 2(1 + 2 + 3 + ... + 48 + 49 + 50)$

which equals $2 \times \frac{50(50 + 1)}{2} = 50(50 + 1) = 50 \times 51 = 2550$.

$$\sum_{k=1}^{20}(4k + 12) = 4(\sum_{k=1}^{20} k) + \sum_{k=1}^{20} 12$$

$$= 4 \times \frac{20(20 + 1)}{2} + 12\times20 = 840 + 240 = 1080$$

For the last part, the minimum possible sum for 7 consecutive numbers is 1

+ 2 + 3 + 4 + 5 + 6 + 7 = 28 and the minimum for 8 is 1 + 2 + 3 + 4 + 5 + 6

+ 7 + 8 = 36. Thus, there must be at most 7 consecutive numbers. We try:

2 consecutive numbers: a + a + 1 = 32 -> 2a = 31 -> no solution.

3 consecutive numbers: a + a + 1 + a + 2 = 32 -> 3a = 29 -> no

solution.

4 consecutive numbers: a + a + 1 + a + 2 + a + 3 = 32 -> 4a = 26 ->

no solution.

5 consecutive numbers: a + a + 1 + a + 2 + a + 3 + a + 4 = 32 -> 5a = 22

-> no solution.

6 consecutive numbers: a + a + 1 + a + 2 + a + 3 + a + 4 + a + 5 = 32 ->

6a = 17 -> no solution.

7 consecutive numbers: a + a + 1 + a + 2 + a + 3 + a + 4 + a + 5 + a + 6 =

32 -> 7a = 11 -> no solution.

Thus, there is no set of positive integers that sums to 32.

Josh derived the sum for an arithmetic progression with first term $a$, difference $d$ and $n$ terms, and used it to solve the remaining questions:

To prove the sum formula we can do the following:

$S_n = a + (a + d) + (a + 2d) + ... + (a + [n-2]d) + (a + [n-1]d)$

If we reverse the sum we get:

$S_n = (a + [n-1]d) + (a + [n-2]d) + ... + (a + 2d) + (a + d) + a$

Note that there are n terms in both these expressions. Adding the two

equations together, matching up the terms from the first sequence with

their reversed terms gives us

$2S_n = (2a + [n-1]d) + (2a + [n-1]d) + ... + (2a + [n-1]d)$

Note that there are still n terms in this expression. Since they are all

now the same, we now actually have n(2a + [n-1]d) so all that is left is to

divide by two to give:

$S_n = (\frac{n}{2})(2a + [n-1]d)$.**After how many terms would $17+21+25+”¦$ be greater than $1000$?**

We can use the formula backwards to find $n$. We can

see that $a = 17$ and $d = 4$ and that $S_n > 1000$

Therefore

$\frac{n}{2}(34 + 4[n-1]) > 1000$

$30n + 4n^2 > 2000$

$2n^2 + 15n - 1000 > 0$

$n > ( - 15 + \sqrt{225 + 8000} ) / 4$

$n > 18.922946...$

$n = 19$ (I only took the +ve solution of n since we can't have a negative number of terms)**Can you find the sum of all the integers less than 1000 which are not divisible by 2 or 3?**

We have to consider a way to express the numbers which aren't divisible by two or three.

The first few are: $1, 5, 7, 11, 13, 17, 19, 23, 25...$

We can see that we have repeating differences of $4$ and $2$, but if we

separate them up we can get two sequences with differences of $6$:

$, 7, 13, 19, 25...$ and $5, 11, 17, 23...$

For the first sequence, $a = 1, d = 6$ and the $n^{th}$ term is the biggest number

of the form $6m + 1$ that is still less than $1000$. I found this number to be

$997$.

$1 + 6(n-1) = 997$

$n - 1 = 166$

$n = 167$

So $S_167 = 83.5(2 + 996) = 83333$

For the second sequence I did the same thing, but with $a = 5$ and the $n^{th}$

term being the largest number of the form $6m + 5$ but still less than $1000$,

which is $995$.

$5 + 6(n-1) = 995$

$(n-1) = 165$

$n = 166$

So $S_166 = 83(10 + 990) = 83000$

Therefore the sum of all integers that are not divisible by $2$ or $3$ and are

less than $1000$ is $83000 + 83333 = 166,333$**Can you find a set of consecutive positive integers whose sum is 32?**

Consecutive => d = 1

So $\frac{n}{2}(2a + n - 1) = 32$, so $n(2a + n - 1) = 64$

Since n > 0, I decided to work with increasingly large values of n up

until the point where they began to return smaller and smaller decimals or

negatives or any other sign that I wouldn't get any more valid values of

a.

$n = 1$

$2a = 64$

$a = 32$ <- this is only one number though so it doesn't really count!

$n = 2$

$4a + 2 = 64$

$4a = 62$

$a = 15.5$ <- invalid number since the series consists of positive integers

$n = 3$

$6a + 6 = 64$

$6a = 58$

$a = 9.66...$ <- invalid number since the series consists of positive integers

At this point I realised a quicker way to get $a$: rearrange the equation to

make a the subject.

$n(2a + n - 1) = 64$

$2a + n - 1 = 64/n$ [1]

$2a = 64/n + 1 - n = (64 + n - n^2)/n$

$a = (64 + n - n^2)/2n$

By the line marked [1] I could see that n had to be even since otherwise we

would get a non-integer value for 64/n which would lead to a non-integer

a.

$n = 4$ gives $a = 6.5$ Not valid

$n = 6$ gives $a = 2.8333$ Not valid

$n = 8$ gives $a = 0.5$ Not valid

$n = 10$ gives $a = -1.3$ Not valid

As n -> infinity, the constant term in the numerator and

the n term in the numerator become arbitrarily small compared to the -n^2.

So a therefore approaches (-n^2)/2n = -n/2 and so a will continue getting

more and more negative and so we can stop checking n terms and conclude

that there are no sets of consecutive positive integers whose sum is 32.

Rajeev had a different approach for the penultimate question:**Can you find the sum of all the integers less than 1000 which are not divisible by 2 or 3?**

Here we first find the sum of all integers which are divisible by 2 and then find the sum of all integers which are divisible by 3 and add these two up and then subtract the sum of the integers that are divisible by 6

$999 \times 1000 /2 = 499500$

Multiples of 2:

$499 \times 1000 /2 = 249500$

Multiples of 3:

$333 \times 1002/2 = 166833$

As multiples of 2 and 3 have duplications they must be removed by counting the number of multiples of 6:

$166 x 1002/2 = 83166$

$499500 - (249500 +166833 - 83166)= 166333$

The sum of all the integers less than 1000 which are not divisible by 2 or 3 is166333

Well done to Abi from Wilberforce Sixth Form College who sent in a clear solution similar to Josh's and Amrit's.

## Teachers' Resources

### Why do this problem?

This problem provides an introduction to summing arithmetic series, and allows students to discover for themselves the formulae used to calculate such sums. By seeing a particular case, students can perceive the structure and see where the general method for summing such series comes from.

The problem could be used to introduce $\sum$ notation.

### Possible approach

You may wish to show the video, in which Alison works out $\sum_{i=1}^{10} i$ in silence, or you may wish to recreate the video for yourself on the board.

Then write up $\sum_{i=1}^{100} i$ and ask students to adapt Alison's method to work it out. Share answers and explanations of how they worked it out.

Next, give students the following questions:

$2+4+6+\dots+96+98+100$

$\sum_{k=1}^{20} (4k+12)$

$37+42+47+52+\dots+102+107+112$

"Can you adapt the method to work out these three sums? In a while I'm going to give you another question like these and you'll need to be able 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.

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

"Imagine a sequence that starts at $a$ and goes up in equal steps to the $n^{th}$ term which is $l$. Can you use what you did with the numerical examples to create a formula for the sum of the series?"

Give students time to think and discuss in pairs and then share their suggestions.

"What if you were asked to find the sum of the first $n$ terms of the sequence $a, (a+d), (a + 2d), (a + 3d) $ and so on - can you adapt your formula?"

Again, allow some time for discussion before bringing the class together to share what they did.

Finally, "Can you use your formula to work out after how many terms would $17+21+25+\dots$ be greater than $1000$?"

### Key questions

What can you say about the sum of the first and last, and the second and penultimate terms of an arithmetic sequence?

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

### Possible extension

Challenge students to find the sum of all the integers less than $1000$ which are not divisible by $2$ or $3$.

Summats Clear would make a nice extension challenge for students who have found this problem straightforward.

### Possible support

Slick Summing explores the same content as this problem but introduces new ideas more slowly and does not use $\sum$ notation.