Abstract.We present a geometric interpretation of Pascal’s formula for sums of powers of integers and extend the interpretation to the formula for sums of powers of arithmetic progres- sions. Related interpretations of a few other formulas are also discussed.
1. INTRODUCTION. In the sixteenth and seventeenth centuries, several mathe- maticians came up with different generalizable derivations of the formula for sums of powers of integers, but Blaise Pascal (1623–1662) was the first mathematician who
http://dx.doi.org/10.4169/amer.math.monthly.119.01.058 MSC: Primary 11B57, Secondary 11B25
explicitly stated the general formula by applying thetelescoping-sumtechnique to the following binomial expansion (see, for example, [2] and [4, pp. 82–83]):
(i+1)p+1−ip+1= Xp k=0
p+1 p−k
ip−k.
Summing the equality fromi =1 tonand expanding the binomial coefficients p+1p−k yields
(n+1)p+1−1= Xp
k=0
(p+1)ã ã ã(p−k+1) (k+1)!
Xn i=1
ip−k.
Dividing byp+1 and writing the products as binomial coefficients yields (n+1)p+1−1
p+1 =
Xp k=0
pã ã ã(p−k+1) (k+1)!
Xn i=1
ip−k
= Xp
k=0 p k
k+1
Xn i=1
ip−k.
Notice that the coefficient ofPn
i=1ipis 1. Writing the equation as an explicit formula for sums of powers of integers yields
Xn i=1
ip = (n+1)p+1
p+1 − n+1 p+1+
Xp−1 k=1
p k
k+1
Xn i=1
ip−k
!
. (1)
To the best of our knowledge, only the arithmetic series, not the series of powered components of an arithmetic sequence, have been interpreted geometrically (see, for example, [3, p. 35] and [1, p. 14]). In this note, we provide a geometric interpretation of formula (1) starting from p=1, followed by interpretations of related formulas.
Our interpretation is also applicable to sums of powers of arithmetic progressions.
2. SUMS OF INTEGERS AND SUMS OF SQUARES. Figure 1 (cf. [5]) shows that the total area (the number) of all the squares is equal to the difference between the area of the enclosing right triangle and the areas of all the small right triangles. That is,
Xn i=1
i = (n+1)2
2 −n+1 2 ,
which is equation (1) when p=1. Similarly, Figure 2, the three-dimensional exten- sion of Figure 1, illustrates that the total volume of all the cubes can be obtained by subtracting the excess from the enclosing right-angle square pyramid. That is,
Xn i=1
i2= (n+1)3
3 − n+1
3 +2
2 Xn
i=1
i
! , which is equation (1) whenp=2.
January 2012] NOTES 59
n
n n+1
n+1
=
− ì (n +1)
Figure 1. Subtracting the excess from the enclosing triangle yields the sum of integers.
n
n n
=
n+1
n+1 n+1
− ì(n+1)
− ì2Pn i=1i
Figure 2. Subtracting the excess from the enclosing pyramid yields the sum of squares.
3. SUMS OF CUBES. The left-hand side of Figure 2 is a step pyramid whose ith layer (counting down from top) containsi ìi unit cubes, forming a square-base box of unit height. This step pyramid is enclosed by the pyramid on the right-hand side, which has an(n+1)ì(n+1) square base and comes to a point at the top,n+1 units above the base. By analogy, extending the step pyramid to the fourth dimension yields a hyper step pyramid whoseith layer is a unit-height hyper box with aniìiìi cubical base. This hyper step pyramid is enclosed by a hyper pyramid that has an(n+ 1)ì(n+1)ì(n+1)cubical base and a point at the top. To facilitate our discussion, let the orientation of this enclosing hyper pyramid be our frame of reference for the termsbaseandtop.
Whenp=2, the basic pieces whose volumes have to be subtracted off come in two shapes, triangular prisms and small pyramids. Both shapes have unit-square bases (2- dimensional), but the tops of triangular prisms are unit-length edges (1-dimensional) and the tops of small pyramids are points (0-dimensional). When p= 3, the basic pieces whose hyper volumes have to be subtracted off will come in three shapes: all have unit cubical bases and their tops will be unit squares, unit-length edges, or points.
LetWpkdenote a basic piece whose base and top are of dimensionspandkrespectively.
For example, a small pyramid and a triangular prism in Figure 2 can be denoted byW20 andW21. Note that the hyper volume of aWkp is 1/(|p−k| +1)(it can be found by, for example, integration).
In Figure 2, if we turn a step pyramid with annìnbase into one with an(n+1)ì (n+1)base by adding unit cubes to each square layer along two adjacent sides of the
(a) (b)
Figure 3. The base (a) and the top (b) of the third-layer L-shaped wedge.
square and adding a unit cube at the top, the intersections of the enclosing pyramid with these added cubes in a single layer would form an L-shaped wedge whose base and top are depicted in Figure 3. By analogy, when p =3, unit hyper cubes would have to be added along three adjacent sides of each cubical layer and at the top to make a hyper step pyramid one unit larger along each edge, and it is the intersections of the enclosing hyper pyramid with these added hyper cubes that give theW3ks. The union of theW3ks from a single layer yields a region whose base is a three-sided wall of a hollow cube with unit thickness (see Figure 4(a)) and whose top is the inside surface of a similar wall (see Figure 4(b)). Looking at the base of this region from theith layer, one would seei2bases ofW32s along each side of the wall. Thus there are 3Pn
i=1i2 W32s. Similarly, along each edge of this wall arei bases of W31s and there are three edges per wall (see Figure 4) for the total of 3Pn
i=1i W31s. In addition, there is one W30at the vertex of each layer and at the top. So the equation becomes
Xn i=1
i3 = (n+1)4
4 − n+1
4 +3
2 Xn
i=1
i2+3 3
Xn i=1
i
! , which is equation (1) whenp=3.
(a) (b)
Figure 4. The base (a) and the top (b) of the third-layer excess region. (The dotted lines are meant to be virtual and employed to provideperspectiveto the recessed corner made by the three mutually perpendicular planes.)
4. THE GENERAL EXTENSION. In order to determine the number of hyper edges for each hyper wall in higher dimensions, note that the wall in Figure 4(b) contain 3 (1-dimensional) edges because 2 out of 3 (2-dimensional) sides are needed to form an edge, for the total of 32
edges withi1 W31s along each edge. For a general case of Pn
i=1ip, k out of p (p−1)-dimensional sides are required to form a (p−k)- dimensional edge for the total of pk
such edges with ip−k Wpp−ks along each edge.
January 2012] NOTES 61
Thus the total hyper volume of allWpp−ks is
p k
k+1
Xn i=1
ip−k, k =1, . . . ,p−1.
In addition, all(p−1)-dimensional sides meet at the vertex of each layer to form a (p+1)-dimensional small pyramid. A similar excess can be found at the top of the (p+1)-dimensional enclosing pyramid for the total ofn+1 small hyper pyramids.
Subtracting the excess from the enclosing hyper pyramid yields equation (1).
5. A SIMILAR FORMULA WITH ALTERNATING SIGNS. It is straightforward to verify that applying the telescoping-sum technique to the binomial expansion of ip+1−(i −1)p+1yields the following formula for sums of powers of integers:
Xn i=1
ip = np+1 p+1+
Xp k=1
(−1)k+1 kp k+1
Xn i=1
ip−k. (2)
This formula is more appealing to some because the first term on the right is a power ofn, notn+1. Its geometric interpretation is very similar to that of Pascal’s formula, only slightly more complicated. Figure 5 (see also [6] and [1, p. 20]) illustrates the dif- ferences between the two interpretations whenp=2. Instead of subtracting triangular prisms from the enclosing pyramid, we need to add inverted prisms to the inscribed pyramid. However, where two edges meet at a corner, the two inverted prisms overlap to form an inverted pyramid, which has to be subtracted off. These inverted prisms are the parts of unit cubes in the step pyramid that lie outside the inscribed pyramid. In summary, looking at a single layer, instead of subtracting the union of all nonoverlap- pingW2ks from the enclosing pyramid, we have to add the union of allW12s, some of which overlap, to the inscribed pyramid.
= + ì2Xni=1i
− ìn
Figure 5. Adding the excess (and subtracting the overlaps) to the inscribed pyramid yields the sum of squares.
Analogously, whenp=3, Figure 4(a) would depict the top of the union of allW23s from thefourthlayer, and the corresponding base would be similar to itsoutsidesur- face. For each unit along the edges of the wall, twoW23s overlap to form aW13, which has to be subtracted off. At the vertex of each layer, threeW23s and threeW13s overlap (see Figure 4), nullifying each other. Thus their overlap,W03, has to be added back.
A third-level overlap (in yet a higher dimension) involves 41
Wpp−1s, 42
Wpp−2s, and
4 3
Wpp−3s with alternating signs. So, we would have already added 4−6+4= 2 copies of the overlap and thus have to subtract oneWpp−4. In general, a jth-level over- lap involvesPj
i=1(−1)i+1 j+1i
Wpp−is. Since the alternating-sign sum of all binomial
coefficients from an expansion is zero (which can be verified by expanding(x−x)n), the sum of the numbers ofWpp−is is 2 for odd js and 0 for even js, which means that we have to alternately subtract and add consecutive levels of overlaps, in order to retain exactly one copy, resulting in equation (2).
For both interpretations, the (hyper) step pyramids can also be made symmetrical (so that the centers of all the layers are vertically aligned) if only for aesthetics. When p =2 (see Figures 2 and 5), such symmetrization would result in twice the number of triangular prisms, each with half the volume of an original prism, and four times the number of small pyramids, each with one-fourth the volume of an original small pyramid. In higher dimensions, analogous symmetrization would result inWkps and Wkps being evenly split among opposite sides, edges, or corners, without changing their total hyper volumes.
6. SUMS OF POWERS OF ARITHMETIC PROGRESSIONS. Our geometric interpretation can be applied to Pascal’s formula for sums of powers of arithmetic progressions;a,a+d, . . . ,a+(n−1)d. For arithmetic series, the width of the first column of the steps in Figure 1 is replaced byawhile that of other columns is replaced byd. The height of each layer is still one. Thus the enclosing triangle has widtha+nd and height ad +n. Figure 6 illustrates the interpretation when p=2. It is straightfor- ward to verify that extending this construction to higher dimensions yields the equation
n−1
X
i=0
(a+id)p = (a+nd)p+1 (p+1)d −
"
ap+1 (p+1)d +
Xp k=1
p k
dk k+1
n−1
X
i=0
(a+id)p−k
# .
n
a d d d a d d d= (a+nd)/d
a+nd a+nd
− a/d
a a
−
d dìn
−2Xni=0−1
d a+id Figure 6. The intepretation of sums of squares of arithmetic progressions.
7. COMPLETING THE HYPER CUBE. In [7] and [8], Turner demonstrated a geometrical interpretation of the power-sum formula,
Xn i=1
ip=np+1−
p−1
X
k=0
p k
Xn−1 i=1
ik+1,
for the cases of sums of integers and sums of squares. Although the equation is an implicit one, her interpretation is so ingenious that it is worth extending to higher dimensions. The interpretation of sums of squares depends on the ability to form a cube
January 2012] NOTES 63
from three step pyramids, two of which contain one layer less than that of the intended- sum pyramid (B and C in Figure 7), and one set of steps which is also one layer shorter. The shortened step pyramids are rotated and placed on top of the intended- sum pyramid to leave room along the diagonal of thenìnìn cube in which the shortened steps can be disassembled to fit (D in Figure 7). In higher dimensions, this can be achieved by successively placing ever-smaller objects at successive levels of edges. Thus, the interpretation of the term pk
is exactly the same as ours while other terms are more or less self-explanatory.
(I)
↑A
←D
←B
C→
⇒
(II) (i)
B A C
D
A (ii) C B
D
(iii) A
B
C D
Figure 7. (I) Component parts (A, B, C, and D) that form the cube presented from three viewpoints (i), (ii), and (iii) in (II).
ACKNOWLEDGMENTS.We would like to thank anonymous referees for their valuable suggestions that help clarify this note. Remaining errors, if any, are our responsibility.
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//dx.doi.org/10.2307/2691265.
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Institute for Innovative Learning, Mahidol University, Salaya, Nakhon Pathom 73170, Thailand.
pl one@hotmail.com
Multidisciplinary Unit, Faculty of Science, Mahidol University, Ratchathewi, Bangkok 10400, Thailand.
scbpn@mahidol.ac.th