We provide a combinatorial proof of a formula for the sum of evenly spaced binomial coefficients. This identity, along with a generalization, are proved by counting weighted walks on a graph

Benjamin, Arthur T.

Chen, Bob, \u2710

Kindred, Kimberly

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Claremont CollegesScholarship @ ClaremontAll HMC Faculty Publications and Research HMC Faculty Scholarship12-1-2010Sums of Evenly Spaced Binomial CoefficientsArthur T. BenjaminHarvey Mudd CollegeBob Chen '10Harvey Mudd CollegeKimberly KindredHarvey Mudd CollegeThis Article is brought to you for free and open access by the HMC Faculty Scholarship at Scholarship @ Claremont. It has been accepted for inclusionin All HMC Faculty Publications and Research by an authorized administrator of Scholarship @ Claremont. For more information, please contactscholarship@cuc.claremont.edu.Recommended CitationBenjamin, Arthur T., Bob Chen and Kimberly Kindred. "Sums of Evenly Spaced Binomial Coefficients." Mathematics Magazine , Vol.83, No. 5, pp. 370-373, December 2010.Sums of Evenly Spaced Binomial CoefficientsAuthor(s): ARTHUR T. BENJAMIN, BOB CHEN, KIMBERLY KINDREDSource: Mathematics Magazine, Vol. 83, No. 5 (December 2010), pp. 370-373Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/10.4169/002557010X529860 .Accessed: 11/06/2013 19:05Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp .JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact support@jstor.org. .Mathematical Association of America is collaborating with JSTOR to digitize, preserve and extend access toMathematics Magazine.http://www.jstor.org This content downloaded from 134.173.219.117 on Tue, 11 Jun 2013 19:05:03 PMAll use subject to JSTOR Terms and Conditions370 MATHEMATICS MAGAZINESums of Evenly Spaced Binomial CoefficientsARTHUR T. BENJAMINHarvey Mudd CollegeClaremont, CA 91711benjamin@hmc.eduBOB CHENHarvey Mudd CollegeClaremont, CA 91711b2chen@ucsd.eduK IMBERLY KINDREDHarvey Mudd CollegeClaremont, CA 91711kindred@math.hmc.eduTwo of the first identities encountered in a discrete mathematics course are the follow-ing finite sums of binomial coefficients. For n ≥ 0,∑k≥0(nk)= 2n (1)and for n ≥ 1,∑k≥0(n2k)= 2n−1. (2)The sums are finite since(nk) = 0 when k > n. Both of these identities have ele-mentary combinatorial proofs. But when r ≥ 3, the sum∑k≥0 ( nrk) is rarely mentionedbecause its closed form is more complex. (See Gould [1]. A special case appears in [3]as problem 1.42(f).)THEOREM 1. For n ≥ 0 and r ≥ 1,∑k≥0(nrk)= 1rr−1∑j=0(1 + ω j )n, (3)where ω = ei2π/r is a primitive r th root of unity.Notice that when r = 1 or 2, we have ω = 1 or −1, respectively, and the formulasin equations (1) and (2) are directly obtained. When r = 3, we haveω = ei2π/3 = −1 +√3 i2and then Theorem 1 yields, for n ≥ 0,∑k≥0(n3k)= 2n + m3,Math. Mag. 83 (2010) 370–373. doi:10.4169/002557010X529860. c© Mathematical Association of AmericaThis content downloaded from 134.173.219.117 on Tue, 11 Jun 2013 19:05:03 PMAll use subject to JSTOR Terms and ConditionsVOL. 83, NO. 5, DECEMBER 2010 371where m depends on n and is equal to 2, 1,−1,−2,−1, 1, when n is congruent to0, 1, 2, 3, 4, 5 (mod 6), respectively. Likewise when r = 4, we have ω = i , and we get∑k≥0(n4k)= 2n + m2n/24,where m = 2, 1, 0, −1, −2, −1, 0, 1, when n ≥ 1 is congruent, respectively, to0, 1, 2, 3, 4, 5, 6, 7 (mod 8). (When n = 0, this formula needs to be adjusted, since00 = 1.)A generalization of Theorem 1 (which appears in Gould [1] in modified form) alsohas an attractive closed form.THEOREM 2. For any integers 0 ≤ a < r and n ≥ 0,∑k≥0(na + rk)= 1rr−1∑j=0ω− ja(1 + ω j )n, (4)where ω = ei2π/r is a primitive r th root of unity.While Theorems 1 and 2 have succinct algebraic explanations using the binomialtheorem (see [2], [3]), our goal is to prove them combinatorially. In a combinato-rial proof, an identity is proved by counting a problem in two different ways. Ourproofs will utilize the graph Cr , the directed, looped cycle graph with vertex setV = {0, 1, . . . , r − 1} such that for each vertex j , there is an arc to vertex j andj + 1 (mod r ). (See Figure 1.) We define an n-walk to be a walk on Cr that takesexactly n steps. A walk that begins and ends at the same vertex is said to be closed;otherwise it is open. For example, when r = 5, the walk 3, 4, 4, 0, 1, 1, 1, 2 is an open7-walk. It makes 4 forward moves and 3 stationary moves. Another way to describethis walk would beX = (x0; x1, x2, x3, x4, x5, x6, x7) = (3; F, S, F, F, S, S, F)where x0 indicates the initial vertex and the other values of xi indicate whether the i thstep is forward or stationary. Clearly, an n-walk that begins at x0 and makes m forwardmoves will end up at vertex x0 + m (mod r ).3 2104Figure 1 The looped cycle graph C5.This content downloaded from 134.173.219.117 on Tue, 11 Jun 2013 19:05:03 PMAll use subject to JSTOR Terms and Conditions372 MATHEMATICS MAGAZINECombinatorial Proof of Theorem 1QUESTION. How many closed n-walks on Cr begin at vertex 0?Answer 1. Starting at vertex 0, there are(nm)n-walks that take m forward steps. Fora walk to be closed, m must be a multiple of r . Consequently, our first answer is simply∑k≥0(nkr).Answer 2. First we observe that there are as many closed n-walks that start at vertex0 as start at vertex 1 or 2 or . . . or r − 1. Thus it suffices to prove that the total numberof closed n-walks on Cr is∑r−1j=0(1 + ω j )n. We accomplish this by assigning eachn-walk (whether it be open or closed) a weight that will depend on its initial vertexand the number of forward moves. Specifically, an n-walk with initial vertex x0 = jthat makes m forward moves will be assigned a weight of ω jm . The 7-walk on C5 inthe previous example has j = 3, and m = 4 and therefore has weight ω12 = ω2 sinceω5 = 1. Equivalently, a walk that begins at vertex j and ends at vertex j + m (mod r )has weight ω jm. In particular, any closed walk will have weight ω0 = 1.Another way to think of the weight of a walk beginning at vertex j is that eachstationary step is given weight 1 and each forward step in the walk is given weightω j , and the weight of the walk is defined as the product of the weights of its steps.Consequently, the total weight of all n-walks that begin at j is (1 + ω j)n , since each(1 + ω j) represents a choice in our walk to make a stationary or forward move. (Alter-natively, (1 + ω j )n = ∑k≥0 (nk)ω j k is the sum of the weights of all n-walks starting atj since (nk)ω j k is the total weight of all such walks with k forward steps.) Summingover all possible starting points,r−1∑j=0(1 + ω j )n (5)counts the total weight of all n-walks (open and closed) on Cr .Our goal is to show that (5) counts the total number of all closed n-walks on Cr .Since each closed walk has weight 1, it suffices to show that the total weight of allopen walks is zero. Consider an open walk X0 that begins at vertex 0 and ends atvertex m = 0. Then X0 generates the orbit {X0, X1, . . . , Xr−1} where walk X j startsat vertex j , and then follows the same forward and stationary instructions as X0, endingat vertex j + m (mod r ), with weight ω jm . Summing a finite geometric series, the totalweight of the n-walks in this orbit isr−1∑j=0ω jm = 1 − ωmr1 − ωm = 0,since ωr = 1 and ωm = 1. Since every open walk appears in exactly one orbit, eachwith total weight zero, the total weight of all open walks is zero, as desired. Summa-rizing, for walks on Cr ,the number of closed n-walks = the total weight of all closed n-walks= the total weight of all n-walks=r−1∑j=0(1 + ω j )n.Hence, the number of closed n-walks that begin at 0 is 1r∑r−1j=0(1 + ω j )n .This content downloaded from 134.173.219.117 on Tue, 11 Jun 2013 19:05:03 PMAll use subject to JSTOR Terms and ConditionsVOL. 83, NO. 5, DECEMBER 2010 373Combinatorial Proof of Theorem 2In this proof, an n-walk on Cr that starts at vertex j and makes m forward movesis defined to have weight ω− jaωmj = ω(m−a) j . Hence any walk that makes a + rkforward moves has weight ωrk j = 1. Just like in the proof of Theorem 1, the totalweight of all n-walks on Cr is∑r−1j=0 ω− ja(1 + ω j )n. The theorem follows since thewalks that make forward progress m = a + rk can be placed into orbits of total weight∑r−1j=0 ω(m−a) j = 0.Theorems 1 and 2 can also be expressed in terms of trigonometric functions [1],sometimes without mentioning any complex numbers. Suppose ν = eiπ/r is a primitive2r th root of unity so that ν2 = ω. Then using Euler’s formula, e−iθ + eiθ = 2 cos θ , wemay write the summand as(1 + ω j )n = [v j (v− j + v j )]n = vnj (e−iπ j/r + eiπ j/r)n = vnj(2 cosπ jr)n.In particular, if n is a multiple of r , say n = qr , then∑k≥0(nrk)= 2nrr−1∑j=0(−1)q j(cosπ jr)n(6)can be expressed entirely with real numbers. This is the form presented in [1]. Like-wise, Theorem 2 simplifies to the same right hand side of (6) when n = qr + 2a.Where do we go from here? A natural problem might be to try to count walks onother graphs to discover other identities. Conversely, we hope this technique may allowus to combinatorially understand other identities that mix binomial coefficients withcomplex numbers. For example, Identity 2.24 in [1] says for r > 1,∑k≥11(krr) = r−1∑k=1−ωk(1 − ωk)r−1 log 1 − ωk−ωk ,where ω is a primitive r th root of unity. Perhaps with the right combinatorial perspec-tive, this identity will not appear nearly so complex after all.Acknowledgment The authors thank Keith Dsouza and Michael Krebs, whose talk at the Fall 2009 meetingof the Southern California and Nevada Section of the MAA provided the motivation for this work.REFERENCES1. Henry W. Gould, Combinatorial Identities, Morgantown Printing and Binding, Morgantown, WV, 1972.2. David R. Guichard, Sums of selected binomial coefficients, College Math. J. 26(3) (1995) 209–213. doi:10.2307/26873453. La´szlo´ Lova´sz, Combinatorial Problems and Exercises, 2nd ed., American Mathematical Society, Providence,RI, 2007.Summary We provide a combinatorial proof of a formula for the sum of evenly spaced binomial coefficients,∑k≥0(nrk). This identity, along with a generalization, are proved by counting weighted walks on a graph.This content downloaded from 134.173.219.117 on Tue, 11 Jun 2013 19:05:03 PMAll use subject to JSTOR Terms and Conditions

Sums of Evenly Spaced Binomial Coefficients

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Sums of Evenly Spaced Binomial Coefficients

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