Math 409/409GHistory of Mathematics

Sums of Squares and Cubes

In the lesson on figurative numbers you saw how the triangular and square numbers were used to find formulas for sums of counting numbers, even counting numbers, and odd counting numbers.

( 1)1 2 3

2

n nn

2 4 6 2 ( 1)n n n

21 3 5 (2 1)n n

In this lesson we will use what we have learned about the triangular and square numbers to derive formulas for the sums of squares and cubes.

One of these derivations will be geometric and the other will be algebraic.

2 2 2 21 2 3 ?n

3 3 3 31 2 3 ?n

But first, lets review a bit.

The triangular numbers are generated by the iterative relation .

4t 5 4 5t t

1n nt t n

1

2

3

4

5

6

7

8

1

3

6

10

15

21

28

36

t

t

t

t

t

t

t

t

In the lesson on the sum of triangular numbers we derived the formula for the sum of the first n triangular numbers.

The square numbers are 2.ns n

1 2

( 1)( 2)

6n

n n nt t t

To find a formula for the sum of the squares of the first n counting numbers, let’s first consider the sum

Use a diagram of the first four square numbers to represent this sum.

2 2 2 21 2 3 ?n

2 2 2 21 2 3 4 .

Add enough dots to form a rectangular array.

Do you see any pattern in the dots we added?

The added dots can be counted using the triangular numbers.

1

2

3

4

5

6

7

8

1

3

6

10

15

21

28

36

t

t

t

t

t

t

t

t

The number of dots in this array is

This number is also the product of the (horizontal) length and (vertical) width of the array.

2 2 2 21 2 32 .1 3 4 t t t

Clearly the length is 10 and the width is 4. But we want to generalize our formula for

to

We need to express that 10 in terms of 4.

2 2 2 21 2 3 4 2 2 2 21 2 3 .n

But 10 is the forth triangular number.

So let’s add another row of dots at the top.

1

2

3

4

5

6

7

8

1

3

6

10

15

21

28

36

t

t

t

t

t

t

t

t

The number of dots in the array is now

And the length of the array is t4 and its width is 4 + 1.

1 2 3 42 2 2 21 2 3 4 .t t t t

This gives

And generalizing this gives us

2 2 2 24 1 2 3 4(4 1) (1 2 3 4 ) ( )t t t t t

2 2 21 2( 1) (1 2 ) ( )n nn t n t t t

We know that

and

Plugging these into our last equation gives

2 2 21 2

2 2 2

( 1) (1 2 ) ( )

( 1) ( 1)( 2)( 1) (1 2 )

2 6

n nn t n t t t

n n n n nn n

( 1)

2n

n nt

1 2

( 1)( 2)

6n

n n nt t t

And then solving for

gives us the formula we were looking for.

2 2 2

2 2

2 2

2

2

( 1) ( 1)( 2)( 1) (1 2 )

2 6

( 1)(1 2 ) 3( 1) (

( 1)(2 1)1

2

26

)6

n

n

n n n nn n

n nn n

n

n

nn

2 2 2 21 2 3 n

Do you see a pattern in the successive values of

3

3 3

3 3 3

3 3 3 3

3 3 3 3 3

1

1 2

1 2 3

1 2 3 4

1

9

36

100

21 2 3 4 5 25

3 3 3 31 2 3 ?n

3 3 3 31 2 3 ?n

Each sum is a perfect square.

Do you see a pattern in the numbers that are squared?

3 2

3 3 2

3 3 3 2

3 3 3 3 2

3 3 3 3 3 2

1 1

1 2 9

1 2 3 36

1

1

3

6

10

1

2 3 4 100

1 2 3 4 5 2 525

They are the triangular numbers. 1

2

3

4

5

6

7

8

1

3

6

10

15

21

28

36

t

t

t

t

t

t

t

t

3 2 21

3 3 2 22

3 3 3 2 23

3 3 3 3 2 24

3 3 3 3 3 2 25

1 1

1 2 3

1 2 3 6

1 2 3 4 10

1 2 3 4 5 15

t

t

t

t

t

So in general,

And since , we now have a

formula for the sum of the cubes of the first n counting numbers. Namely,

3 3 3 3 21 2 3 .nn t

( 1)

2n

n nt

23 3 3 3 ( 1)1 2 3

2

n nn

This ends the lesson on

Sums of Squares and Cubes