Proofs without Words 61

6.7 A Fibonacci identity

Consider next the traditional Fibonacci sequence, 0, 1, 1, 2, 3, 5, 8, 13, 21.... Each

number is the sum of the previous two. If f

n

denotes the n th term in this sequence, so

that f

0

= 0, f

1

= 1, and f

n

+ f

n+1

= f

n+2

, then this diagram is offered as proof that

f

2

0

+ f

2

1

+ ···+ f

2

n

= f

n

f

n+1

:

1 1

2

3

5

8

13

21

Do you see it? The squares arise from the Fibonacci numbers; their arrangement exactly

ensures the Fibonacci recursion: f

n

+ f

n+1

= f

n+2

. Since the squares altogether at stage n

correspond to a rectangle with sides f

n

by f

n+1

, we deduce the desired identity.

6.8 A sum of cubes

Consider next the following diagram, aiming to prove the cubic sum identity:

5

4

3

2

1

1

3

+ 2

3

+ ···+ n

3

= (1 + 2 + ···+ n)

2

Can you see how it works? Note that we have one tiny gold square of size 1. If we put

the two red rectangles together, it makes in total two red squares of size 2. And similarly

there are three orange squares of size 3, four squares of size 4, and five squares of size 5.

So the sum of the areas of all the colored squares is the sum of cubes 1

3

+ 2

3

+ 3

3

+ ···+ n

3

(why?). On the other hand, they are assembled into a large square, whose side length is

1 + 2 + 3 + ···+ n. So the two sides of the desired identity both represent the total area of

the large square.