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A look back at 2014, and forward to 2015

December 30, 2014 7 comments

This time last year I wrote this post reviewing 2013 and looking to 2014. In the summer I looked again at it here, and discussed my year so far. Now, as the year draws to a close and I am reading all these #Nurture1415 posts it feels like a good time to reflect again.

My 2014

At home

I’ve had a good 2014, I’ve spent some great time with my family, watched my daughter grow from a baby/toddler into a real little person and seen a lot of other family.

Studies and the blog

I’ve continued to work on my masters, and to write this blog. Both of which have helped me improve as,a teacher, and both of which have been enjoyable. There has been a higher proportion of maths puzzles finding there way onto the blog this year. This hasn’t been a conscious decision, but I have really enjoyed working on them.

Maths

I’ve managed to read a few more maths books, and I have managed to get deeper into topology, as I had hoped to this year. I’ve also delved deeper into group theory, another old favourite of mine, and I particularly enjoyed exploring the tests of divisibility.

Teaching

When I wrote the post last year, I thought I’d be in the job I was in for a long time to come. In reality I’d decided to move on and found another job by the end of February (I think). This was a massive change. I moved jobs, schools and authorities and there were some real challenges. I had a full set of new classes to build relationships with, and a full set of new colleagues to get to know. The new school is similar on many levels to the old one, but it is also infinitely different too. I feel I’ve joined a great team, and that I’ve already made some great friends amongst my colleagues. I feel that with most classes I’ve built up decent relationships and am making progress, and I feel I’m getting to grips with the new role.

CPD

I’ve been on some great CPD events this year. I’m on a Teaching Leaders course, I attended ResearchEd York, Northern Rocks, Maths Conference 2014 and teachmeets (I even presented at one!) The key messages for me is challenge everything, don’t just accept anything, ensure there’s something to back it up, and even then don’t just assumed it will work in all contexts.

Education in 2014

2014 will forever be remembered for that day in July when the news that Gove had gone shocked the nation. I wrote about my feelings at the time and you can read them here. I can’t say I’ve seen much difference in policy since he left, and I feel the move was made purely to detoxify the brand in the run up to the election.

It was also the year we got to see the draft maths A Level curriculum, which looks good, but not radically different, and the approved specifications for the new maths GCSE. I’m excited about the new GCSE as I think it addresses many if the short comings that the current one has, although I’d have liked to see calculus and Heron’s Formula make an appearance.

The Sutton Trust released a report in 2014 entitled ‘what makes great teaching”, it was my favourite type if report, one that backs up the things I thought with plenty of evidence. The crux of its finding being “great teaching is that which leads to great progress“. You can download the report in full here free of charge.

It also saw the first teaching of the new “Core Maths” suite of post 16 qualifications. We are a pilot school, and I’m quite excited by the prospect, although it’s not been without teething problems so far.

Hopes for 2015

Last year I hoped that the new curriculum would increase the rigour of the maths being taught and that it helps prepare learners for A Level. I still hope this, although I realise now it is a longer term hope. As is the hope that the new GCSE system will eliminate the threshold pass and the gaming we have seen with early entry and other such things. And I think it’s too early to tell if the new routes into teaching can bring down the high turnover we experience.

I still hope that the inherent inequalities present in the UK education system, and wider society, can be addressed.

I hope to find more time to spend with my family, to read and investigate further areas of maths this year.

I hope to continue to improve my practice and to get better at my job.

I hope to see an end to the ridiculous pseudo-context “real life” problems we often see in exams.

And I hope to make a real difference to the learners I’m in front of in 2015, to increase their maths knowledge and skill but also their respect for, and love of, mathematics. A number of my Y13 learners have applied for maths degree courses, and I hope they enjoy them.

I hope you have enjoyed reading this post, and have had a great year, and festive period, yourselves. Here’s hoping we all have a happy new year, and a fantastic 2015

A nice area puzzle

December 29, 2014 1 comment

At some point over the last few days Danny Brown (@dannytybrown) tweeted this lovely puzzle out:

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While waiting in the car this afternoon I saw the screenshot I’d taken of it and had a think about it. Initially I’d made a daft mistake with the base of a triangle and got the wrong answer (5:6 if you were wondering), but after a rethink and some in head workings I got to an answer I’m now happy with. I have since written the solution down and want to write about it here.

Firstly, when attacking problems like this I like to sketch them out. This step is one many students are reluctant to take, and I try to drill it into them that sketching is always helpful in understanding problems. If I’d had a pen and paper handy when I’d attacked this problem originally I’d have sketched it and not made my daft error.

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From the sketch it is easy to see the area of shape A, this was easy to visualise.

In my working I then sketched shape B:

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When I visualised this I made my daft error, which was to assume the triangle had base x. From the sketch it’s easy to see that it’s not, and it will need calculating, for this I used Pythagoras’s Theorem:

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To calculate the area of the triangle I’d need an angle, so I thought about what I knew, the angle at E is made from a fold along EF so the angle must be equal to DEF.

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Using the tan ration of the right triangle I got tan (theta) to be rt3.

Then it was a case of calculating the height and then the area:

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The working written out properly looks a lot more thorough than my phone jottings:

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This is a nice puzzle that should be accessible to higher GCSE students and definitely A Level Students, but I worry that most would give up. We need to be giving our students the tools to unlock this sort of problem. I’m not sure how we can do that explicitly. I set these tasks, give them hints and then walk them through my thinking to model how I would attack them, and this has a great effect for many. But I wonder if this is enough.

Perimeter, the Hero’s way

December 28, 2014 2 comments

Forgive the title, but I do love Heron’s Formula, (named after Hero of Alexandria) it’s my favourite geometric formula and this is only the second time I’ve had need to mention it in a blog post. The first was Area the Hero’s way.

This post is a look at my solution to this beautiful puzzle which was set by Ed Southall (@solvemymaths) the other day:

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This struck me as a tricky one. The fact it’s a right angled triangle meant my brain was crying out “we’ve got to use trigonometry”, but I couldn’t see an obvious was to do it. We have the area of the triangle, and there is an incircle in the picture so I figured we’d need to use the relationship that the area of a triangle is equal to the semiperimeter multiplied by the radius of the incircle. It was possibly the fact that this relationship mentions semiperimeter that put Heron’s Formula in my mind.

I sketched the problem and filled in what lengths I could deduce using circle theorems, filling gaps with variables:

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From this the semiperimeter was nice to work out, so I went down Hero’s route.

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Which led me to:

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I had one equation, but two variables. I needed another equation in the same two variables, so I used the aforementioned relationship from incircle:

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I rearranged this for x (as it was the simplest rearrangement) and subbed it into the other equation:

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I then solved this equation for r, discounting the negative, as a radius can’t be negative. Once I had r I could use the relationship area/radius = semiperimeter so twice area/radius =perimeter:

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Which rounds to 44 cm. (The question asks to the nearest cm.)

I love this question, and I’m happy with my solution. I think it’s fairly elegant and uses a nice array of mathematics, but I can’t help but think that I’ve missed something blinding obvious that would have led to a simpler solution. If you solved it a different way, I’d love to hear how you did it. (I’d also like to hear if you did it the same way!) I asked Ed how he solved it, and he used the fact that the point where the incircle meets the longest side of a triangle splits the longest side into two segments, the product of whose lengths is the area. This is a nice relationship, and not one I’d known, so I will look to explore it. This could have saved be a bit if working, as I could have jumped to tge bit where I wrote x=6.

Rectangle Puzzle

December 27, 2014 Leave a comment

At some point over the last few days Danny Brown (@dannytybrown) tweeted this puzzle:

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The picture shows a unit square with two congruent rectangles and a number of triangle. I looked,at the puzzle, saw no obvious answer, realised it would need a bit of thought so saved the photo until I found some time.

Yesterday I had some time, so I started by sketching the puzzle and filling in things I knew:

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I quickly realised that the angles for all the triangle were the same, and thus they were all similar. (For some reason I missed one of the triangles on this page.) I couldn’t work out how I could find a missing length. I worked out the length if the diagonal of the rectangle using Pythagoras’s Theorem but it didn’t really help.

Then I drew some perpendiculars from each end of the diagonal and had a breakthrough.

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If two rectangles share a diagonal, and a side length, then they must be congruent. This meant that I now knew the length of the bottom was 2x + 2y, as it’s a unit square then 2x + 2y = 1 so x+y=1/2.

I filled in these lengths:

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Redrew my similar triangles and solved:

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As no one likes an irrational denominator I rationalised as ended up with the satisfying solution 2-rt3.

I love this puzzle, and think it could be used as enrichment for high ability learners of many ages. After I’d solved it I looked at some solutions others had posted, and was thankful to see that other people also unnecessarily use trigonometry, and I was amazed I’d managed to avoid it, given my track record!

Venn Diagrams, Algebra and the New GCSE

December 17, 2014 15 comments

I’m a fan of Venn Diagrams (and the more versatile, lesser know Euler Diagrams), and the uses they have when dealing with probability. I was glad to see then installed in their full glory to the new GCSE curriculum. They are an excellent way of displaying data, and they can give rise to some great questions which test a range of mathematical skills in a different (sometimes unfamiliar) context. While I was looking for the question mentioned in this post I came across this question:

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It’s from the AQA higher SAMs and I absolutely love it. It looks simple from the outset, but it actually covers a number of topics. You need to have an understanding of Venn Diagrams, you need an understanding of Probability Theory and you need to be proficient in Algebra.

Algebra underpins everything in mathematics, and the biggest flaw in the current GCSE is that it allows people to gain a food pass without Algebraic proficiency. I think we need to see more questions like this, that mix topics and place algebra at the heart of the question, to ensure people leave the GCSE course with the skills needed to progress further in mathematics.

The question

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I couldn’t help but explore it. My first thought was, “I need to find x”, this was fairly easily achieved using the fact that there are 120 coins. It’s just a case if forming and solving the equation.

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Discounting the negative x, of course. Then it was just a case of substituting the values in and finding the probability from the Venn Diagram.

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A lovely problem.

“Next Level” question

December 15, 2014 3 comments

Back in November I wrote this post outlining the changes to the new GCSE maths curriculum and included some questions I had enjoyed from the SAMs and some that I had thought were less good.

One of my Y13 students had read the post and had had a go at some of the questions, he came to me in school and told me that he’d seen a “next level” question on my blog, that it had taken him ages to realise how to do it and that he couldn’t believe it was on the new foundation GCSE. I asked him which question it was and he said it was this one:

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I considered the question. A shows a sector of a circle with an angle of pi/2 radians. I should say 90 degrees I suppose, as it’s a GCSE question. B shows the same square but this time there are 4 sectors, each with the same angle, again pi/2 radians 90 degrees. There are no lengths marked on.

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This question is relatively straightforward once you pick up the key piece of information, that the squares are the same. You don’t even need to use the angles to work out the sectors as we are dealing with as they are quarter circles. You assign a value for side length and work through each one, showing the areas are the same.

I think it’s a great question, and I love the fact that these two shading arrangements give the same area. I did, however, wonder how future foundation students would cope with it given that a Y13, who scored a good A in maths AS and As in his other 3 ASs in Y12, referred to it as “Next Level”.

When discussing it with said student it turned out that the maths wasn’t the issue, it was that there was no lengths marked on, and he hadn’t realised straight away that you could assign a variable to work though.

I assume this is down to the nature of the current examinations, and their tendency to give scaffolding all the way through. I like the rigour of the new qualifications, and the fact they are designed to build mathematical thinking all the way through, but I fear that for some teachers ensuring this is built in will take some getting used to.

Find the radius puzzle

December 13, 2014 1 comment

Yesterday I came across this photo on my phone:

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It’s a question that I found on a wall in one of the classrooms at school when I started in September and thought, “that’s a nice puzzle”, then promptly forgot about. When I found it, I thought “let’s explore this one.”

I sketched it out and had a look, my brain made its usually first stop at trigonometry.

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After spending a while deriving (accidentally) that cos x = adj/hyp for right angled triangles using the sine rule, various trigonometric identities and Pythagoras’s Theorem I decided there must be a better way. I sketched the problem on a coordinate grid so the centre of the circle was at (r,0) when r is the radius (ie so the y axis is a tangent) and looked at the equation:

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As I know the dimensions of the rectangle, I can deduce that the point (2,r-1) is on the circumference, sub these values in and solve for r.

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When I have r= 1 or 5 I can discount 1 as I know r>2, thus the radius is 5 cm . I was surprised by this answer, as it felt like the answer should be 4, not sure why.

Bizarrely, as I was working through this Jo Morgan (@mathsjem) posted this which included this similar puzzle:

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So I used the same method to solve that:

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I was happy with the solution, but I had a real feeling that I was missing something obvious that would lead to a much more concise solution, so I sketched again, this time I dropped a perpendicular from the point on the circumference which meets the rectangle:

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A right angled triangle, with side lengths r-2, r-1, r. It’s only the most famous RAT of all, a 3,4,5 triangle!

I followed the algebra:

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A much more concise method. This method made me think Chris Smith’s (@aap03102) puzzle (the one Jo shared) would be better to use in class, as it’s much more likely that students will know a 3,4,5 triangle than a 20,21,29 triangle!

Tests of divisibility

December 11, 2014 Leave a comment

Today I was marking my year 9 classes books and came across some work on prime factor decomposition and tests of divisibility. Yesterday I had been arguing with Colin Beveridge (@icecolbeveridge) about the use of formula triangles, (see the comment section on this post) and any other method of anything that was presented in the fashion “do it this way and don’t worry about the why.” I felt a wave of hypocrisy flow over me. I had never explained the tests of divisibility to the class, and furthermore I didn’t even know myself why they worked! I thought I’d explore them and see what I came up with.

If the last digit is divisible by 2, the number is divisible by 2.

This test of divisibility is obvious, the definition of an even number is that it has 2 as a prime factor, and all even numbers have a last number divisible by 2. No further exploration needed. I know why this one works and I’m certain the class do too.

If the sum if the digits is a multiple of 3, then the number is divisible by 3.

This is a fact I’ve known since I was at primary school back in the 1980s, but I can honestly say that I’ve never thought about why it works. I wrote a couple if equations out, realised I should be working in modulo 3 and came up with this:

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Which basically boils down to the fact any number can be split into the digits multiplied by a powerful of ten.

A= (x0)10^0 + (x1)10^1 +… + (xn)10^n

As (10^n) mod 3 = 1 for all natural numbers (and 0) then it follows that:

(A)mod 3 = (x0 +x1+…+xn) mod 3

Which implies our test of divisibility. This also implies the test if divisibility for 9 (ie is the sum of digits is divisible by 9 then so is the number) as (10^n)mod9 = 1 for all natural numbers and 0. To prove you just follow the exact working but in mod 9.

A number is divisible by 4 if it’s last two digits are divisible by 4.

This one uses the fact that if two numbers are both divisible by a number then so is their sum.

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We know any number bigger than 100 can be expressed 100a + b where a and b are natural numberso and b<100. We also know that as 4 is a factor of 100, 100a is divisible by 4 if a is a natural number. Hence the whole number is divisible by 4 iff the best is divisible by 4, and b is the last two digits.

From this we can deduce the test of divisibility for 8 (ie a number larger than 1000 is divisible by 8 iff the last three digits are divisible by 8) as 1000 is divisible by 8. The proof is the same, but you split the number into 1000a + b where a and b are natural numbers and b is less than 1000.

Testing for 5 and 6

Testing for 5 (the last number is 5 or 0) is another obvious one that needs no further exploration. And the test for 6 is simple, if 2 and 3 are prime factors 6 must be a factor, as all numbers can be expressed as a product of their prime factors and 6 is the product of 2 and 3.

Testing divisibility for 10

Again, this is obvious as we are talking about a base ten system! That leaves just one more test.

To test for divisibility by 7, take the last digit, double it. Take this away from the number you are left with if you subtract the last number and divide by ten. If the answer is 0 or divisible by 7 (ie if the answer mod 7 = 0) then the original number us divisible by 7.

What a ridiculously complicated divisibility test. It looks much nicer algebraically:

(10a + b) mod 7 = 0 (a and b are natural numbers, b < 10)

iff

(a -2b)mod7 = 0

Once I expressed it algebraically I had two simultaneous equations and solved like this:

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I enjoyed working through them and now I’m confident I know why these tests work, but I’m fairly sure that the ones that aren’t obvious would be too complicated to explain to year 9. This made me think about their use, and I certainly think it’s fine to use them even without understanding. Which brought me back to the formula triangles debate. This is one that will rage, but as I explained in my post on it, I don’t have a great issue with people who understand the algebra using them. I also don’t have a great issue with people who struggle at maths and aren’t going to pursue it past GCSE using them. My issue is with higher attainers who could and should understand the algebra being taught then with no deeper conceptual understanding. I guess it’s a topic I will need to think more on.

Area of a semi-circle – the return

December 8, 2014 3 comments

Last week I wrote this piece in which I relayed my thought process when solving this problem from Ed Southall (@solvemymaths).

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After sharing it a lot of people tweeted me their solutions and I thought I’d look at a few here. I think the variety of solutions people used, and the amount of people who engaged with this puzzle is testament both to how interesting a puzzle this is and also how fantastic maths is. If you did it a different way, do let me know!

My solution was fairly long compared to many of the others. I think this maybe down to my heavy preference to Trigonometry, and my brain’s insistence on using it if possible! Here are some other, more concise methods:

Using the Tangent Ratio

In my solution I used right angled triangle trigonometry to find the lengths of the chords, then used Pythagoras’s Theorem to find the diameter. Steve Atkinson (@Small_Ears) also used trig, spotted that as tan a = 2 realised that tan b had to be 2 and thus the remainder of the diameter must be 12. A much more efficient and concise method, but not as much fun!

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Steve’s method made me realise we are dealing with three similar triangles, and that fact alone is enough to know the ratio of the sides are equal, and hence the rest if the diameter must be 12. This idea was the one used by Rob Rolfe (@robrolfemaths).

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Drawing a line from the centre

Jo Morgan (@mathsjem) tweeted this solution:

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Which I think is brilliant. Instead of calculating the rest of the diameter, Jo went for the rest of the radius, using a line from the centre to the point the perpendicular touches the circumference. Pythagoras’s Theorem and algebraic manipulation drop the radius out quickly implying the answer in great time.

The I only need Pythagoras’s Theorem method

This was tweeted to me by Mr Draper (@mrdrapermaths):

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As you can see, he has used Pythagoras’s Theorem on all three triangles and solved the simultaneous equations to find the necessary information. Possibly not the most concise method, but certainly no less concise than my approach, and one that has a lovely feel to it.

Constructing the geometric mean

Martin Noone (@letsgetmathing) tweeted this fantastic solution:

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It is, I believe, derived from an old method for constructing square roots that I think dates back to ancient Greece. This rule is illustrated in hospital top sketch, then he’s enlarged it to ensure the one is a three. He later tweeted the proof, which he uses as an exercise in lessons and one I may start to use too!

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Ed Southall then tweeted this pic which shows the general rule and how a semi-circle can be used to construct the geometric mean of two numbers, which itself could be used to find the diameter.

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And is a great visual proof that the geometric mean can never exceed arithmetic mean.

Coordinate geometry

Stuart Price (@sxpmaths) informed me that he’d enjoyed the puzzle and left it in his board for his students to complete, and that they used a variety of methods. He mentioned coordinate geometry and equations of circles, so I thought I’d have a crack at that method:

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I drew the axes so that the origin was at the point on the diameter where the perpendicular touches it. Ie 3 cm in from the circumference. I considered the general equation of a circle:

(x-a)^2 + (y-b)^2 = r^2

Then subbed b=0 in as the centre lies on the x axis.

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I know two points on the circumference, (-3,0) and (0,6) so I subbed these in to get two simultaneous equations which I then solved.

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Once I knew a (the x coordinate of the centre) I could work out the radius, and hence the area. A brilliant solution.

Calculus

Stuart also mentioned calculus, specifically implicit differentiation. I’m not sure how this would help, to be honest, and I’d love to see this solution. I did, however, consider integration. I rearranged the equation of the circle:

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I started to integrate but realised I’d need multiple substitutions, so I ran it on Wolfram Alpha instead.

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An nice, but incredibly long winded solution!

Nice summation puzzle

December 6, 2014 Leave a comment

Chris Smith (@aap03102) runs a weekly maths newsletter during termtime. If you’re not on the list, I’d advise you to get on it (drop him a line in twitter). It has some great stuff in it, such as this fraction:

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How ace is that?

He also has a puzzle of the week feature. This week’s looked cool, so I had a crack at it:

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I thought about it a little, and saw that it boiled down to this:

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I had the general term: 1/((n^1/2)+(n+1)^1/2) but I wasn’t sure where that would get me. I tried rationalising the denominator of the first two terms:

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Now I could see where I was getting to, I thought I’d check the general term, to be sure:

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Then it was a case of looking at the sum:

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The sum of (n+1)^1/2 -n^1/2 for n=1 to n=91808

Then I cancelled:

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Leaving just

91809^1/2 -1^1/2

Which is

303-1 =302

A lovely neat solution to a lovely summation problem which I hope to try in my sixth formers soon, and which I hope will work well with the new GCSE specifications.

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