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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.

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