Picture Story
Problem
Getting Started
Student Solutions
Teachers' Resources
Secondary Curriculum Linked
Picture Story
Age 14 to 16
Challenge Level

Jack and Ashley from Sir Harry Smith Community College sent us the following thoughts:

(1+2+3+...+n
)2
is equal to
13
+
23
+
33
+...+
n3
, which can also be written
(

r=1
n
r)
2
=
r=1
n
r
3


We tested this formula with two examples:
13
+
23
+...+
63
=441
, and
1+2+...+6=21,
212
=441
.
13
+
23
+...+
103
=3025
, and
1+2+...+10=55,
552
=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
52
), 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:
52
+4×
52
=(
52
)(1+4)=(
52
)(5)=
5
3

As this is true for the first, second, third...
n
th
level of the area of the
n
th
level is
n3
.
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
(
13
)+(
23
)+(
33
)...+(n1
)3
+(
n3
)
.
So it can be said
(
13
)+(
23
)+(
33
)...+(n1
)3
+(
n3
)​=​(​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
13
is equal to
12
, 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
(
13
+
23
...
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.

Copyright © 1997 - 2021. University of Cambridge. All rights reserved.
NRICH is part of the family of activities in the Millennium Mathematics Project.