**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:

1^{3} = 1^{2}

2^{3} = (1+2)^{2} – 1^{2}

3^{3} = (1+2+3)^{2} – (1+2)^{2}

4^{3} = (1+2+3+4)^{2} – (1+2+3)^{2}

n^{3} = (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} + n^{2} + 2n(1+2+3+4+…+ (n-1)) – (1+2+3+…+ (n-1))^{2}

= n^{2} + 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 + (n^{2} – 2n + 1)/2

Therefore:

2n(1+2+3+4+…+ (n-1)) = 2n ( (n-1)/2 + (n^{2} – 2n + 1)/2 )

2n(1+2+3+4+…+ (n-1)) = n^{2} -n + n^{3} – 2n^{2} + n

Therefore

n^{2} + 2n(1+2+3+4+…+ (n-1)) = n^{2} + n^{2} -n + n^{3} – 2n^{2} + n

n^{2} + 2n(1+2+3+4+…+ (n-1)) = n^{3}

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

**An alternative relationship**

1^{3} = 1^{2}

1^{3}+2^{3} = (1+2)^{2}

1^{3}+2^{3}+3^{3} = (1+2+3)^{2}

1^{3}+2^{3}+3^{3}+…n^{3} = (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:

1^{3}+2^{3}+3^{3}+…k^{3} = (1+2+3+…+k)^{2}

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

ie.

1^{3}+2^{3}+3^{3}+… k^{3} + (k+1)^{3}= (1+2+3+…+k + (k+1))^{2}

LHS:

1^{3}+2^{3}+3^{3}+… k^{3} + (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]^{2 }+ (k+1)^{2} + 2(k+1)(k/2 (1+k)) (sum of a geometric formula)

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

= [1+2+3+…+k]^{2 } + k^{3}+ 3k^{2} + 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.