Picture Story
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Picture Story
Age 14 to 16
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Jack and Ashley from Sir Harry Smith Community College sent us the following thoughts:

(1+2+3+...+n
 ) 2
is equal to
 1 3
+
 2 3
+
 3 3
+...+
 n 3
, which can also be written
 (∑r=1nr) 2
=
r=1
n
r
3

We tested this formula with two examples:
 1 3
+
 2 3
+...+
 6 3
=441
, and
1+2+...+6=21,
 21 2
=441
.
 1 3
+
 2 3
+...+
 10 3
=3025
, and
1+2+...+10=55,
 55 2
=3025
.

An anonymous solver from Mearns Castle School explained very clearly how the diagram shows the two parts of the formula:
The area of the square can be looked at in different ways.
The columns are
1
then
2
then
3
then
4
units long. It is a square, so the area is the length squared which is
(1+2+3+4+5+6
 ) 2
.

Now look at each of the 6 backward L shapes.
We have the indicated square on the diagonal (for example, on the blue L shape
 5 2
), then we have other congruent squares (shaded in the same colour).
Note that in even levels, there are two half-squares which make up one full square. So the area of the blue shaded level is:
 5 2
+4×
 5 2
=(
 5 2
)(1+4)=(
 5 2
)(5)=
5
3

As this is true for the first, second, third...
n
th
level of the area of the
n
th
level is
 n 3
.
This means if there are
n
levels, the length of one side of the square is
1+2+3...+(n1)+n
and thus the area equals
(​1​+​2​+​3...​+​(​n​−​1​)​+​n
 ) 2
.
But considering the area in terms of each level gives
(
 1 3
)+(
 2 3
)+(
 3 3
)...+(n1
 ) 3
+(
 n 3
)
.
So it can be said
(
 1 3
)+(
 2 3
)+(
 3 3
)...+(n1
 ) 3
+(
 n 3
)​=​(​1​+​2​+​3...​+​(​n​−​1​)​+​n
 ) 2
.

John from Takapuna Grammar School used induction to prove the formula:

Using induction we should first prove that
 1 3
is equal to
 1 2
, which is obvious, as they're both equal to
1
.
Now to prove that any added number
(n+1)
would keep the equation satisfied:
The formula for any triangular number
(1+2+3...+n)
is
n(n+1)
2
.

Now we observe the way in which new tiles are put on. There are
(n+1)
squares to be added, which contain
(n+1
 ) 2
units each, making
(n+1
 ) 3
.
One of these squares is placed diagonally to the original square and the other
n
squares are placed along the sides.
Since there are two sides to place against, this means that there are
n
2
squares per side, and because each square is
(n+1)
units long that means there are
n(n+1)
2
units covered on each side.

There are
n(n+1)
2
units on each side of the original square so that means all of the units on each side are covered, leaving the last
(n+1
 ) 2
square (the one that is diagonal to the original square) to make a new square itself.
Finally, we know that each side of this new square is the
 (n+1 )th
triangular number, because the previous square's length was the
n
th
triangular number, and this length has increased by (n+1).
Therefore the equation
(
 1 3
+
 2 3
...
n
3
)=(1+2...+n
 ) 2
holds for all positive whole numbers, because it is true for n=1 and if it's true for any number n then it's true for n + 1.