Sequence Investigation

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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]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  + 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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