# Show that the sum of the cubes of the first n natural numbers equal

```Problem #1: Show that the sum of the cubes of the first n natural numbers equal the
square of the sum of the first n natural numbers.
Solution Path: It would be at good start to show the development of the sum of the first n
natural numbers. We will use the familiar method, supposedly made famous by Gauss.
Sn  1 2  3  ...  n  2 n 1 n


S  n  n 1 n  2 ...  3  2  1
 2S  n  1 n  1 n  1 ...  n  1 n  1 n  1
 2S  nn  1
n
n
n

n n  1

 
S

 n

2


Now we want to show the sum of the cubes of the first n natural numbers.
n  1  n  4n  6n  4n  1
n  n 1  4n 1  6n 1  4n 1 1
n 1  n  2  4n  2  6n  2  4n  2 1
4

4
3
4
4
4
2
3
2
4
3
2
34  2 4  23  42  62  42 1
3
2
2 4 14  13  4
1  6
1  4
1 1
2
If we sum up all of the equations in the above list, we can represent the sum of cubes in
terms of n.
4
4  
4
4 
4

 
n  1  n 4  n 4  n 1  n 1  n  2  ...  34  2 4  2 4 14  n  1 14 







3
2
3
2
2
2

 

 3
  3

4n 3  6n 2  4n  1  4n 1  6n 1  4n 1 1 4n  2  6n  2  4n  2 1 ...  42  62  42 1 4
1  6
1  4
1  1

 


 

3
3
3
3
2
2
2
2




4n 3  n 1  n  2  ...  2  
1  6n 2  n 1  n  2  ...  2  
1  4 n  n 1 n  2 ...  2  1  1 1 1 ...  1 1

 








4Sn 3  6Sn 2  4Sn  S1

Expanding the binomial for (n + 1)4 and making the appropriate substitutions for the sum
of the first n natural number and the first n squared natural numbers, we will derive our
equation for the sum of the first n cubed natural numbers.
n  1 1
4
4
 4Sn 3  6Sn 2  4Sn  S1
nn  12n  1 nn  1
 4

n  4n  6n  4n  11  4Sn 3  6

  2  n
6

 

4
3
2



n 4  4n 3  6n 2  4n  4Sn 3  2n 3  3n 2  n  2n 2  2n  n
n 4  2n 3  n 2  4Sn 3 
n 2 n  1
2
 Sn 3
4
If we take our expression for the sum of the first n natural numbers and square the
expression, we should get this last expression.
nn  1

Sn 
2
n n  12 n 2 n  12
2
Sn    2   4  Sn3


Thus the square of the first n natural numbers equals the sum of the cubes of the first n
natural numbers.

Problem #2: Develop a sum for the first four powers of the first n natural numbers.
Solution Path: We have shown the sum of the first n natural numbers and the sum of the
cubes of the first n natural numbers in an earlier problem. At this moment, we will find
the formula for the sum of the squares of the first n natural numbers and the formula for
the sum of the fourth powers of the first n natural numbers, with the hopes of finding a
closed form for the sum of any natural number power of the first n natural numbers.
First we will find the sum of the squares of the first n natural numbers. To do
this, we will refer to the squares of natural numbers in terms of the cubes of those natural
numbers.
n  1  n  3n  3n  1
n  n 1  3n 1  3n 1 1
n 1  n  2  3n  2  3n  2 1
3
3
2
3
3
2
3
3
33  23  32  32 1
2
23 13  3
1  3
1 1
2

2
At this point, we will add up these equations together. By doing so, we can refer to the
sum of the squares of the first n natural numbers in terms of the sum of the first n natural
numbers and n.
n  1  n
3
3
 n 3  n 1  n 1  n  2 
3
3
3
 33  23  23 13  n  1 13 
3
3n 2  3n  1 3n 1  3n 1 1 3n  2  3n  2 1 32  32 1 3
1  3
1 1
2
2
2
2
2
2
2 

3n 2  n 1  n  2  ...2  
1  3 n  n 1 n  2



2

 2  1  1 1 1
3Sn 2  3Sn  n

We will use these expressions to solve for the sum of the squares of the first n natural
numbers. We will confirm the result of the sum of the first n natural numbers in a
subsequent problem.
n  1 1
3
3
 3Sn 2  3Sn  n
n n  1

n  3n  3n  11  3Sn 2  3
 2  n


3
2
3
1
n 3  n 2  n  3Sn 2 
2
2
1 3 1 2 1
n  n  n  Sn 2
3
2
6
n n  12n  1
 Sn 2
6
Now we will try to develop a pattern for the sum of the fourth powers of the first n
natural numbers. We will use the same format as determining the sum of the squares.

5
n  1  n5  5n 4  10n3  10n2  5n  1
n 5  n 1  5n 1  10n 1  10n 1  5n 1 1
5
4
3
2
n 1  n  2  5n  2  10n  2  10n  2  5n  2 1
5
5
4
3
35  25  52  102  102  52 1
4
3
2
25 15  5
1  10
1  10
1  5
1 1
4
3
2
Adding the above equations will yield the desired result:

2
 1 1
n  1  n
5
5
 n 5  n 1  n 1  n  2 
5
5
 35  25  25 15  n  1 15 
5
5
5n 4  10n 3  10n 2  5n  1 5n 1  10n 1  10n 1  5n 1 1
4
3
5n  2  10n  2  10n  2  5n  2 1
4
3
2
2
 52  102  102  52 1 5
1  10
1  10
1  5
1 1
4
3
2
4
3
2
4
4
3
3
2
2






5n 4  n 1  n  2  ...  2 4  14  10n 3  n 1  n  2  ...  23  13  10n 2  n 1  n  2  ...  22  12 








5 n  n 1 n  2 ...  2  1  1 1 1
1 1 
5Sn 4  10Sn 3  10Sn 2  5Sn  n 
1
2 
 2
n n  12n  1 n n  1
n n  1 

 5

n  1 1  5Sn4  10 4  10
  2  n
6

 



5
5
n 5  5n 4  10n 3  10n 2  5n 

5
5
5
5
n 5  n 4  n 3  n 2  n  5Sn 4 
2
3
2
3

 
 

5 4
5
5
n  2n 3  n 2  2n 3  3n 2  n  n 2  n  n  5Sn 4 
2
3
2
```