*This is an update of the original post. There were a couple of mistakes in the first one, which are now corrected. And I’ve learned some new maths in the process. I’ve credited those below who helped me. The power of Twitter – Thank you!*

What happens when you take sequences of odd or even numbers and make fractions out of them?

These investigations provide some low threshold, high ceiling (in some cases too high for me!) rich investigations. I have been playing with these and encourage you to do so too!

These tasks have lots of benefits in the classroom if used well:

- Purposeful practise. In continuing the sequence and generating subsequent terms, students will repeatedly practice key skills. But rather than just working down a boring list of questions in a text book, they are practicing with a greater sense of purpose, i.e. to try to spot something else. There is a big range here from addition of consecutive numbers and cancelling down fractions to finding the nth term of a quadratic sequence (and harder).
- At key points stop everyone and get a whole classroom discussion going. Ask students to explain their “noticings”. By verbalising their reasoning pupils can grow their individual mathematical confidence. And it builds a classroom culture where other pupils’ noticings are highly valued. Children realise that we can learn by collaborating and listening to each others’ ideas.
- Always look for patterns and aim to generalise. Ask why this happens, does it happen every time and can we build a proof? This is hard. Sometimes what happens is that the students that aren’t ready to make this leap yet. Often they just continue generating more examples. This isn’t a problem as hopefully they are continuing to benefit from purposeful practice of an underlying mathematical skill.

I really encourage you to do some Mathematics and play with them before using them, but if time is short, here are some notes on each one. I have put them in order of difficulty.

### Sum of Odd Numbers

This is probably the most accessible of the four in terms of getting to a generalisation, although actually proving that algebraically is no mean feat!

Before we even get to the fractions, there is some good discussion to be had on mental methods for adding series of odd numbers and spotting that this generates square numbers:

1+3=4 1+3+5=9 1+3+5+7=8+8=16 1+3+5+7+9=10+10+5=25

To generalise this, we need to know that the *n*th odd number is (2*n*-1). Working from the last term backwards, we can write out the sequence as:

1, 3, 5, ..., 2n-5, 2n-3, 2n-1

By adopting the standard approach to find the sum of an arithmetic series, i.e. adding the first to last, second to second last, etc. we see that we get a whole bunch of “2n”s. How many “2n”s? Well there are n numbers so there must be n/2 pairs. So:

2n × n/2 = n²

Now you can start examining the fractions themselves. There is some good practice here of cancelling down fractions and students will realise quite quickly that they all cancel down to 1/3

At this point you might ask some students to generalise whilst some might prefer to continue generating examples.

The generalisation for the denominator builds on the generalisation for numerator. This time we with start with the odd number after the nth odd number and then add a series of odd numbers. Again think about what the last term would be and work backwards.

2n+1, 2n+3, 2n+5, ..., 2n+(2n-5), 2n+(2n-3), 2n+(2n-1)

By combining first and last, second and second last, etc. we can see we now have pairs of “6n”. How many “6n”s? Again, n/2. So the denominator becomes:

6n × n/2 = 3n²

and:

3n²/n² = 3

### Sum of Even numbers

With this one, cancelling down the fractions doesn’t help.

You end up with a pair of quadratic sequences (now corrected – thanks **@solvemymaths**)

Which neatly cancels down to:

### Product of Even Numbers

This one provides lots of practice in “cancelling down” of fractions. Each time you end up with a unitary fraction (i.e. a numerator of one). But does this always happen? And why?

I made a mistake first time on this so I couldn’t find any pattern in the numbers that formed the denominators. Thanks to @mathforge and @wjhornby for pointing out by error.

So the sequence of denominators is 2, 6, 20, 70,… That’s beyond my knowledge of Integer sequences (I did Engineering, you know, not pure Maths!). But @mathsforge sent me this link to oeis.org. That’s another web-site I’ve learned about through this process!

If I had played around a bit longer with this and thought Factorials! then I might have eventually got here (thanks to @MrMattock for sending me this)

### Product of Odd Numbers

This one again provides some cancelling down practice although you are going to be reaching for the calculator pretty quickly.

I’m struggling to spot any pattern in here (no, the next term doesn’t have a 9 for its numerator…), but there must be something, right?

And this is no bad place to take a discussion with your class.

There *must* be something here to be found. We haven’t found it today. Your maths teacher is finding this very hard. Maybe nobody has ever found it. But if we start off with something so simple there must be a way of generalising it. *Surely…?*

*Postscript*: Again, I got a helpful response from @MrMattock. You can see it here, but don’t spoil it, have a go for yourself. The clue is to look for factorials again and don’t express all fractions in their simplest form. Good luck, but I warn you – it’s not pretty!

For the sum of odd numbers one, I thought of it in terms of mean, sharing out the numbers. So, e.g. (1+3+5)/(7+9+11) becomes (3+3+3)/(9+9+9) by taking 2 from the 5 and 11 and ‘giving it’ to the 1 and 7 respectively. Then the fraction clearly(?) cancels down to 1/3. This works for all of them.

For the product of even numbers, you have a typo which has ruined your sequence; the second term should cancel to 1/6 not 1/4.

I don’t know if it’s the sort of thing you’re looking for, but if you cancel down each term in the product by 2 you get factorials:

(2x4x6x8x10)/(12x14x16x18x20) = (1x2x3x4x5)/(6x7x8x9x10)

= (5!)/(10!/5!)

which in general gives you ((n!)^2)/((2n)!), or the reciprocal of (2n)C(n) [i.e. nCr with n=2n and r=n]

There must be a proof somewhere about why (if?) this is always a unit fraction.

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Thank you your comment. I’ve corrected the mistake and included some links to some other comments in the article. Some great maths here!

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Great Stuff. This week I learned of Collatz’s Conjecture from my grandaughter’s latest edition of the kid’s educational magasine ‘Aquila’.

See

https://www.jasondavies.com/collatz-graph/

School students can have fun by working through some examples. However, avoid starting with 27 – it is a monster. It takes 111 steps ascending to 9332 before homing in on 1

(According to Aquila – I haven’t checked it)

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