Clay Kitchings :: Sum of a Series Problem, posed in class on April 18, 2007 :: EMAT 6600

 

 

S1 = 1 + 2 + 3 + … + n

 

S2 = 12 + 22 + 32 + 42 + … + n2

 

S3 = 13 + 23 + 33 + … + n3.

 

Problem: Show that S3 = S12.

 

 

First, in my judgment, it would be helpful to know some closed formulas for each of the three sums.  The S1 is fairly well-known and I have proven this multiple times.  I will take the liberty to state that S1 may also be written as  .

 

 

Next, I sought out to find a closed formula for the sum of the first n square numbers. (It is worth noting that upon completing this problem, it is not necessary to find a closed formula for this sum.)

 

As a fan of using consecutive differences, I chose this method to locate the degree of polynomial involved (if indeed it is a polynomial) as well as the coefficients of each term.

 

I used Excel to quickly find some sums as well as some consecutive differences:

 

n

n2

sum first n2

Consecutive

diffs 1

consec diffs 2

consec diffs 3

1

1

1

 

 

 

2

4

5

4 = 5 - 1

 

 

3

9

14

9

5 = 9 - 4

 

4

16

30

16

7

2 = 7 - 5

5

25

55

25

9

2

6

36

91

36

11

2

7

49

140

49

13

2

8

64

204

64

15

2

9

81

285

81

17

2

10

100

385

100

19

2

 

 

Since the third-order differences are constant, we can conclude that the sum of the first n square integers is cubic.  We set up a system of equations to find the coefficients a, b, c, and d of .

 

 

 

 

 

Solving the system yields a= 1/3, b = 1/2, c = 1/6, and d=0. I solved this by generating a 4x5 matrix on my TI-84 and using the matrix reduced-row echelon form.

 

Using the chart feature of Excel can also be helpful in helping individuals see the relationship once it is determined to be a polynomial of a particular degree.  The following chart was created within Excel.  (You can “add trendline” from the chart menu.)

 

 

 

Therefore, we have a “closed” form to express the sum of the first n square integers. If f(n) is the nth square integer, then .

 

 

 

Let’s now investigate the next sum – the sum of the first n perfect cubes.  I shall perform the operation the same way I did above using Excel.  We obtain the following chart:

 

n

n3

sum first n3

consec diffs 1

consec diffs 2

consec diffs 3

consec diffs 4

1

1

1

 

 

 

 

2

8

9

8

 

 

 

3

27

36

27

19

 

 

4

64

100

64

37

18

 

5

125

225

125

61

24

6

6

216

441

216

91

30

6

7

343

784

343

127

36

6

8

512

1296

512

169

42

6

9

729

2025

729

217

48

6

10

1000

3025

1000

271

54

6

 

We observe that the fourth order differences are constant, thus implying a quartic polynomial. Setting up a system of equations as we did earlier (except with using five instead of four), we obtain the following:

 

 

If f(n) is the sum of the first n cubes, then

 

 

Now let’s try to verify that S3 = S12. 

 

 

Let’s expand:    (which is what we wanted to show.)