COMPLETING THE CUBE

COMPLETING THE CUBEAuthor(s): BARBARA TURNERSource: The Mathematics Teacher, Vol. 70, No. 1 (JANUARY 1977), pp. 67-70Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/27960708 .

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COMPLETING THE CUBE

Classic summation formulas come to life with elegant geometric models.

By BARBARA TURNER California State University

Long Beach, California 90840

MANY interesting properties of sums of numbers can be more readily discovered and comprehended by the student if there is a geometrical way to visualize them. Com

plex algebraic formulas become more un derstandable when we can literally see why they are true. One example of a formula with an elementary geometrical proof is the formula for the sum of the first positive

integers, ̂ ' ?

i - 1

A new three-dimensional geometrical

proof of the formula for /2 iS ?b tained by the process of "completing the

cube," a three-dimensional analog of a well-known two-dimensional geometrical technique used in deriving the formula for

"Completing the cube" involves a con struction whereby the space occupied by a cube is progressively filled up by rectangular boxes. This construction is il lustrated in figures 2 through 6, and is en

ticing to most students. If the rectangular boxes are of different colors, the derivation will be clearer and more enjoyable for the students.

One derivation of the formula * . n(n + \)

'=?;? l'I ?

depends on the formula -1

(i) ' = "2- '.

which has a neat two-dimensional geome trical proof, illustrated in figure 1 for the

Fig. 1. Completing the square

case = 4. Figure 1 illustrates that 1 + 2+ 3 + 4 = 42 - (i + 2 + 3). (Count the num ber of unshaded boxes, the total number of

boxes, and the number of shaded boxes). More generally, the figure would be an

square, consisting of columns of boxes each. The /th column, containing / unshaded boxes, would be completed by ? i shaded boxes. Thus, the column far

January 1977 67



thest to the right would contain unshaded boxes (and no shaded boxes). The next col umn would contain ? 1 unshaded boxes plus I shaded box. Similarly, the column to its left would have ? 2 unshaded boxes and 2 shaded boxes, and so on. Finally, the column farthest to the left would contain 1 unshaded box and ? 1 shaded boxes. Thus, the truth of equation (1 ) is confirmed

geometrically, since the number of un n

shaded boxes, ', equals the total number

of boxes, 2, minus the number of shaded n-i

boxes, '. The geometrical proof of formula (1) in-,

volves literally "completing the square." "Completing the cube" is an analogous

process in three dimensions, which can be used to establish the formula

/ n-i n-i \

(2) ?/? = ?>-{ ?/ + 2>?)

from which can be derived the following formula for the sum of the squares of the first positive integers:

(3) ? .2 + lX2n + 1)

To complete the cube, one must have a

supply of rectangular boxes of two kinds, called rods and squares. "Rods" should be of dimensions X 1 X 1, and "squares" should be of dimensions ?X?XI (where is any positive integer). The unit of mea surement is immaterial. Rods of dimensions X 1 X 1 are called n-rods, and squares of dimensions X ? X I are called w-squares. A 1-rod and a 1-square are identi cal. The volume of an -rod is n; the volume of an n-square is n2.

If the rods and squares are color-coded

according to size, the numerical

relationships involved in the geometrical completion of the cube become clearer and the construction of the model will be more

enjoyable. Cuisenaire rods and squares are

appropriate for construction of the cube and work extremely well.

We shall illustrate completing the cube and derive equation (2) for the case = 4. The construction is illustrated in figures 2

through 6. On top of a 4-square, place a 3-square, a

2-square, and a 1-square, in that order, in the manner indicated in figure 2.

Fig. 2. Volume = l2 + 22 + 32 + 42

The configuration of squares fills up part of a 4 X 4 X 4 cube. To complete the cube, the remainder of this cube must be filled in with rods and squares.

The next step is to erect an upside-down staircase of rods on top of one side of the

configuration, placing a 1-rod on the 2 square, a 2-rod upright on the 3-square, and a 3-rod upright on the 4-square. The

resulting configuration is illustrated in fig ure 3.

Fig. 3. Volume = l2 + 22 + 32 + 42 +1+2 + 3

68 Mathematics Teacher



At this point, the top of the 2-square is

(partially) covered by two 1-squares. Now,

completely cover the top of the 2-square by

placing on it two more 1-squares, as in

figure 4.

Fig. 4. Volume = Ia + 2* + 32 + 4f + 1 + 2 + 3 + 2 X Is

At this point, a reverse L-shaped region of the top of the 3-square is left uncovered.

Now, completely cover the top of the 3

square by placing upright on its exposed surface two 2-squares, at right angles to

each other, as in figure 5.

Fig. 5. Volume = l1 + 2e + 3e + 4* 4- 1 + 2 + 3 +

2 X 1? + 2 X 2*

At this point, a reverse L-shaped region of the top of the 4-square is still uncovered.

Completely cover the top of the 4-square by

placing upright on its exposed surface two

3-squares, at right angles to each other, as

in figure 6.

Fig. 6.

Volume = 4s Volume = I2 + 22 + 32 + 42

+ l+ 2 + 3 + 2Xl2 + 2X22 + 2X32

The configuration of squares and rods

now completely fills the space of a 4 X 4 X

4 cube. Thus, the volume of the

configuration is 4s. But, the volume is also

equal to the sum of the volumes of the rods

and squares. Thus, the volume equals

l2 + 22 + 32 + 42 + 1 + 2 + 3 + 2 X l2 + 2 X 22 + 2 X 32.

Thus,

'^ + ' + ^48? t-l ?-1 i-l

Hence,

'2 = 43- ( ' + 2'2 ) ?

This formula is the special case of (2), when = 4.

The geometrical process of "completing the cube" is the same for any number of

squares. Hence, for any integer > 1,

completing the cube gives a three-dimen sional proof of the formula (2).

From (2), the formula ^ /* g*ven by

(3) is easily derived.

January 1977 69



In fact, since (2) holds for all integers > 1,

n+i / 2 = ( + -

[ + *2 i ?

1=1 = ?=1 1

Hence,

( + I)2 + 3?/2 = ( + I)3

- /

Thus,

(3) ( + \)(2 + 1)

Thus, completing the cube yields a three dimensional geometrical proof of (3), the well-known formula for the sum of the

squares of the first positive integers. The study of numerical relationships

through the use of geometrical models may

provide insight through which further alge braic identities are discovered without the use of geometry. For example, there is a

resemblance in the structure of equations (1) and (2), the formulas obtained by com

pleting the square and completing the cube; so it is natural to wonder if a similar pat tern continues for sums of kth powers of

integers for k > 2. The following more

general equation, suggested by (1) and (2), can be guessed, and in fact proved.

(4) '*

Equations (1) and (2) are the instances of

(4) for the cases k = 1 and k = 2. The significance of equation (4) is that it

can be used to express the sum of the kth

powers of the first positive integers in terms of the sums of the fh powers of the first positive integers for all j< k. In fact, one can derive the following formula from

equation (4).

(5) A .k _(n+

- (n + 1)*

hl k+l

((: : ??+( k + 1

Thus, a formula expressing ^ as a func

tion of is obtained once the formulas for

* are known for all j < k.

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70 Mathematics Teacher



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COMPLETING THE CUBE