Sequence Investigation

This is a nice investigation idea from Nrich.  The above screen capture is from their Picture Story puzzle.  We have successive cubes – a 1x1x1 cube, a 2x2x2 cube etc.

The cubes are then rearranged to give the following shape.  The puzzle is then to use this information to discover a mathematical relationship.  This was my first attempt at this:

13 = 12
23 = (1+2)2 – 12
33 = (1+2+3)2 – (1+2)2
43 = (1+2+3+4)2 – (1+2+3)2

n3 = (1+2+3+4+…+n)2 – (1+2+3+…+ (n-1))2

This is not an especially attractive relationship – but nevertheless we have discovered a mathematical relationship using the geometrical figures above. Next let’s see why the RHS is the same as the LHS.

RHS:

(1+2+3+4+…+n)2 – (1+2+3+…+ (n-1))2

= ([1+2+3+4+…+ (n-1)] + n)2 – (1+2+3+…+ (n-1))2

= (1+2+3+…+ (n-1))2 + n2 + 2n(1+2+3+4+…+ (n-1)) – (1+2+3+…+ (n-1))2

= n2 + 2n(1+2+3+4+…+ (n-1))

next we notice that 1+2+3+4+…+ (n-1) is the sum of an arithmetic sequence first term 1, common difference 1 so we have:

1+2+3+4+…+ (n-1) = (n-1)/2 (1 + (n-1) )

1+2+3+4+…+ (n-1) = (n-1)/2 + (n-1)2/2

1+2+3+4+…+ (n-1) = (n-1)/2 + (n2 – 2n + 1)/2

Therefore:

2n(1+2+3+4+…+ (n-1)) = 2n ( (n-1)/2 + (n2 – 2n + 1)/2 )

2n(1+2+3+4+…+ (n-1)) = n2 -n + n3 – 2n2 + n

Therefore

n2 + 2n(1+2+3+4+…+ (n-1)) = n2 + n2 -n + n3 – 2n2 + n
n2 + 2n(1+2+3+4+…+ (n-1)) = n3

and we have shown that the RHS does indeed simplify to the LHS – as we would expect.

An alternative relationship

13 = 12
13+23 = (1+2)2
13+23+33  = (1+2+3)2
13+23+33+…n3  = (1+2+3+…+n)2

This looks a bit nicer – and this is a well known relationship between cubes and squares.  Could we prove this using induction?  Well we can show it’s true for n =1.  Then we can assume true for n=k:

13+23+33+…k3  = (1+2+3+…+k)2

Then we want to show true for n = k+1

ie.

13+23+33+… k3  + (k+1)3= (1+2+3+…+k + (k+1))2

LHS:

13+23+33+… k3  + (k+1)3

= (1+2+3+…+k)2 + (k+1)3

RHS:

(1+2+3+…+k + (k+1))2

= ([1+2+3+…+k] + (k+1) )2

= [1+2+3+…+k]2 + (k+1)2 + 2(k+1)[1+2+3+…+k]

= [1+2+3+…+k]+ (k+1)2 + 2(k+1)(k/2 (1+k))   (sum of a geometric formula)

= [1+2+3+…+k]+ (k+1)2 + 2(k+1)(k/2 (1+k))

= [1+2+3+…+k] + k3+ 3k2 + 3k + 1

= (1+2+3+…+k)2 + (k+1)3

Therefore we have shown that the LHS = RHS and using our induction steps have shown it’s true for all n.  (Write this more formally for a real proof question in IB!)

So there we go – a couple of different mathematical relationships derived from a simple geometric pattern – and been able to prove the second one (the first one would proceed in a similar manner).  This sort of free-style pattern investigation where you see what maths you can find in a pattern could make an interesting maths IA topic.

IB Revision

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If you’re already thinking about your coursework then it’s probably also time to start planning some revision, either for the end of Year 12 school exams or Year 13 final exams. There’s a really great website that I would strongly recommend students use – you choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021), and then have the following resources:

Screen Shot 2018-03-19 at 4.42.05 PM.pngThe Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and each area then has a number of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial.  Really useful!

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The Practice Exams section takes you to ready made exams on each topic – again with worked solutions.  This also has some harder exams for those students aiming for 6s and 7s and the Past IB Exams section takes you to full video worked solutions to every question on every past paper – and you can also get a prediction exam for the upcoming year.

I would really recommend everyone making use of this – there is a mixture of a lot of free content as well as premium content so have a look and see what you think.