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Georgia Journal of Science 66 No. 2 Scholarly Contributions from the Article 9 Membership and Others

2008 The aF scinating Mathematical Beauty of Triangular Numbers

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Recommended Citation (2008) "The asF cinating Mathematical Beauty of Triangular Numbers," Georgia Journal of Science, Vol. 66, No. 2, Article 9. Available at: https://digitalcommons.gaacademy.org/gjs/vol66/iss2/9

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rHE TASCINATING MATHEMATICAL BEAUTY OF TRIAhIGUIAR hIUMBERS

Stacy Cobb, Natasha Patterson, Mulatu Lemma Savannah State university Savannah, GA 31404

ABSTRACT The triangular nurnbers ane formed by partial surn of the !+2+3+4+5+6+7 ....+n {21. trn other words, triangular numbers are those counting numbers that can be written BS: 1+2+3+...+

T_1^i T,: 1+2-3 Tr: 1+2+3-6 To:1+2+3+4-10 Tr: 7+2+3+4+5: L5 'l Ttf, +2+3+4+5+6- 27 - ^ ' Tr: 1+2+3+4+ 5+6+7 - 7,8 Tu: 1+2+3+4+5+5+7+8- 36 Tn: 1 + 2+3+4+5+6+ 7 +B+9:45 T,* - L +2+3+4+5+6+7 +8+9+10:55 trn this pap€r we lnvestigate sorne important properties of triangular numbers. Some important results dealing with the mathematical concept of triangular numbers will be proved. We try our best to give short and readable proofs. Idost of the results ane supplemented with €xamples.

Key Words; Trlangular nurnbers, Per{ect , Pascal , and perfect numbers.

INTRODUCTION The sequenc€ 1, 3, 6,10, 15, ...3 n{n + 1!/Z shows up in rnany places of mathematics {1}" The Greek called ther',n triangular numbers {1}. The T, ir a figurate nun'lber that can be represented in the form of a triangular grid o{ points wh ede the first rcw contains a single element and each subsequent row contains one nnore elernent than the previous one as shown below {21. * * ** * ** *** * ** *** **** * ** *8* **** *****

{<* **{< ,,<**r5 ***** ,<8**** O tJ 6 10 15 21

Published by Digital Commons @ the Georgia Academy of Science, 2008 1 Georgia Journal of Science, Vol. 66 [2008], Art. 9 L59

Mathematicians have been fascinated for many years by the properties and patterns of triangular numbers (21. We can easily hunt for triangular numbers using the formula: n(n +L) r":-;, n > o.

The first 20 triangular numbers are as follows.

TB Tu T$ Trc Tt7 T,, T,, Tro T1 T2 T3 T4 T-t T6 T7 T8 Ts Tfi T,, T,,

1 3 6 10 15 21 28 36 45 55 66 78 91 fi5 ua 136 153 171 190 2L0

THE MAIN BESI.JITS Proposition li Every triangular number is a binomial coelficient. Proof without words: Refer to the following Pascal's (2) and see the underlined numbers.

1 11. t27 1331 14647 15101051 161520L561 172135352171 18285670562881 193684126L2684369L

Remarkl: Also note that Tn which is a binomial " z = ("i') coefficient for each n z 1.

Theorem 1: T is a triangular number e 8T+1 is a perfect square-

Proof: (i) (+) Assume T is a triangular number.

n(n+L) l-et-'. T=- t;, n a positive .'

n(n+L) +8T=8--2^

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n( n + 1) =+8T+1-8 1 2

n( n + 1) 2 8T+1_8 ' ' + 2 2

+BT+ 1_ 8n'+8n+ 2 2

2+4n+1) =+87 + 1 - 2(4n

(2n+ 1X2n+ 1)

tZn+1)2

Hence, 8T+ 1 is a perfect square.

(ii) (€) Assurne 8T+1 is aperfect square. Then 8T+1 is odd - for some n:') positive integ er n, w€ have 8T+ 1- (2n+1)2 4n'+4n+1 =+T:n( . - 2 Hence, T is a triangular number.

BV (i) and (ii) the theorem is proved. Example 1. 6 is a triangular number - 8(6)+ 1 :49 is a perf ecl square. Example 2. 8(15)+1:121, a perfect square =+ 15 is a triangular num- ber.

1 Corollary 1. T is a triangular number =+ n : 1S an integer. Proof: The corollary easily follows by Theorem 1.

Theor em 2: If T* and Tn are triangular nurnbers, then

T **n : T * * \ + mn for m and n positive .

Published by Digital Commons @ the Georgia Academy of Science, 2008 3 Georgia Journal of Science, Vol. 66 [2008], Art. 9 161

Proof: m(m + 1) Note,Tm= & Tn : Then 2 ry.

d d m( m+1) n(' n+1) ' ,f + I + mn 22 + +mn 22 m +m+n +n +mn

m' +m+n' +n+Zmn +Zmn+n' + - m' m+n

22 Example 3. Consider T, and Tn. Note that Tr:6 and T n:10. Observe that and Tr*n : T, : 28 and Q + Tn + 3(4) - 6 + 10 + 72 : 28. Hence , T3*n: T. + Tn + 3(4). Theorem 3: If T,n and Tn are triangular numbers, then :TT T n +T.Tm- L n- I

"r"rr, m(m + 7) Note,Tm: & Tn : Then 2 ry. +1) 1) (n T*Tn+ T^tTnt ry{ryJ * {rn -1)m -L)n -m(m2222

f zz z z I f 22 z z I :4L4jlmn +mn +nm +mn I Imn -mn -nm +mn I

2m'n' + Zmn Zmn + 1) mn (mn + 1) 4:T:T {mn

:T mn

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Example 4. L-etm : 6 and n = 7. Then To = 2L andT, : 28. n(n+7) 42 (43\ By using ,,= we get Truxr, = Tor= = 903 7, T We also have TuTu = 15(27\ = 315 and TuT, + TuT, = 588 + 315 = 903.

Hence, Toxrt= Tn : TuT, + Tu + Tu

Lemma 1. The sum of two consecutive triangular numbers is a perfect square.

Proof: Let T n.r and Tn and be any two consecutive triangular numbers, such that

(n n( n-+7) r,, -!)(n) and r" : - 2"2 Then, r- ,'T (n-7)(n),n(n+l) ,<-Trr---n-r"22

n2211 -n +n +n zn 2 2 which is a perfect square.

Example 5. Let Tu and T rbe any consecutive triangular numbers. Then Tu+ Tr:27 + 289 :49, which is a pertect square"

(2k+1) LemmaZ. lz+Zz+Z2+42+ + kz:k(k+1) 6 Proof. We can easily prove the lernma using induction. Example6.Letk:5 Then 12 +22 +32 +42 +52:1 +4 +9 + 16 + 25 :55.

5( 6) (11) Also , wo have 55 and hence 1z + 2z + gz + 4z + 5z: 6 - 5( 6) (1 1) 6

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Theorem 4. If Tube triangular numbers for k > 0, then we have nt .f-_ n(n+l)(n + 2) .L't- e^ 6-

Proof: To prove the theorem, we apply diuide and_conquer method by considering two cds€s: (1) If n is even, Sog n -2k, then T + Tr+ r+... T n:(T, +T r)+(fs +T n)+... +(T zx.r+T ru|

-t 22+42+(2k)' (by tr-emma 1) - 4(l'+22 +kz)

ak(2k +1)/k + 1) i-----j- (by Lemma 2)" 6 n( n+L)(n+2) asn:Zk 6 (2) lf n is odd, say n: 2k* 1, then

f r+ T r+... + T,:(T, +T r)+(Ts +T n)+... +(T zx.r+T ru)+ T run, (2k +1){2k + 2) 22 +42 + -f.- ... +( zk)z * fty Lemma 1 and definition 2 of Tn)

(2k - 4(12+22+...+k') n +1)(2k +2) 2

+t)(k +t) z) - *k(?k n trzk rt {zk-+ ( bv Lemm a z.)

2L(2k +L)(Zk + Z) 3(Zk +il{Zk + Z) 66 o

- (2k +L)(2k +2)(2k +3) 6

n( n +L)(n + 2) - as n - Zk+L 6 BV (1) and (2) the Theorem is proved.

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Example 7 . Let n :5. Then Iq _ Tr+ T r+Tr+ T o+ T u: 5( 6) (x 1+3+6+10+ 15:35. we also h.',,nn='tave , : 35 and hence \.'- 5(6)(7) -U , ,I H k =l b Theorem 5. For any n, the number 1 + 9 + 92 + 93 + + 9" is a triangular number. 9'*'-1 Proof: Let T:1 + 9 + 92+ 93 + + 9n. Then T- . 8 By Theorem 1, it is suffice to prove that 8T+1 is a perfect {uare. We will apply the divide and conquer method as in Theorern 4.

(1) If n is even, SdV n -2k, then

gT+1 _ g (r'.1 -, )+ 1 :9n+ 1- gzk*1: g(grn)-(3r**r)r, which is a perf ect square. 5

(2) 1I n is odd, say n - 2k+1 , then

8T+1 - B (9'.''8 -1 )+ 1 -9n+t*gzk+2-()zk+r)2, which a perfect square. By (1)and {2), the theorem is proved.

Theorern 6. If T,be triangular numbers for n>1, then we have

E+:2 :1 Proof , L; n:1 1, : i2 ?-rn( n+1) : D::,1* \) *l : 2(t):2

Published by Digital Commons @ the Georgia Academy of Science, 2008 7 Georgia Journal of Science, Vol. 66 [2008], Art. 9 165

Proposition Z.The difference of the squares of two consecutive trian- gular numbers is a .

(n -1)n (n +l)n proof: consid er Tn-1: .- and T-n - Z Z Then,: (T,)r-(T,-r),: (ry\-fT u '" )' \ ) \. )

: nn + zn' + n2 - nn - 2n' + n' :nn' : 444 n3 Example 8. Let Q and Trbeany two consecutive triangulur r,,.r*bers. Then (Tr)'- (Tu)' :'282 - 212 : (28 : -21X28 - 2U: {a9)Ul= 73, which is a perfect cube.

Proposition 3: T is a triangular is number =+ 9T+1 is a triangular number.

Proof: Assume T is a triangular number.

LetT n(n+1) - 2 +97 _ 9n(n+1) 2 9n(n + 1) =+ 9T+1 - *t 2 _ 9n(n+1) *2 22 _ 9n(n +L) + 2 2 9n2 +9n+2 2

2 m( m+1) - 2', .wherem:3n+1.

Hence, 9T+1 is a triangular number.

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Example 9.Let T, bn the triangular number. Then 9 Tr+1- 45, which is a triangular number.

Froposition 4: n_zk-t + 2r< + 2k+1 + + 22k-? isatriangularnum- ber.

Proof: Note that n -Zk-| + 2k + + 22k-z * Zk-t (1+2+22+...+2n-')

: Zk-t (Zn _ 1) : zu (zu -\ 2.! : m( m+1) .wherem-2*-1 2 Hence, n is a triangular number

Example 1O. Let k =3. Then n = 23-L + 23 + 2a : 28,which is a tri- angular number:

Proposition 5. n -- l+2+3+4+ ...+{2k - 1) is a triangular number.

Proof: Note that n :2k(2u-t) 2 m(m+1) : --, wnere m = 2k - l. 2 Hence, n is a triangular number

Example 11. Let k : 3. Then 1+2+3+4+5+6+7=28, which is a tri- angular number.

Proposition 6. Every [3] is a triangular number.

Proof: Let n be a perfect number. Then n = 2k-1 (2k - 1) where 2k - 7 zke! m(n!11), is prime [3]. Note that n = 2x-t 12x - 1) = -t) - where 2 2 m = 2k - i. H"r,." n is a triangular number.

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Proposition 7 . Let T, b" a triangular number. Then: 2 * (1)\-/ T-n :T-n + T-n-1 . T-n+1 2 :2*T * 2lTE n-l - n T n-I

Proof: (n+1)(n+2) (1) : n:nt) n- WehaveTn*Tn-r*Tn-1 "( 222*(n-1) : (n +1) + 2n' - n' -2n 2n' + 2n + nn + 2n3 - n' - 2n 24 *no nn +2n3 +n' : S"oltf(')

:l r2

(2) No,jrn"* r,,-t:W

((n -1) n)(n(n +1) ) 2

2((n -1) n) (n( n +1) ) 4 (n-1) n* n( n+1) - Z* 22

L2*T tn-7'n .*T

Conclusion: We discussed a good number of interesting results on triangular numbers. We strongly believe that this study will be a good encour- agement for more deeper investigation into triangular numbers. Triangular numbers are very interesting numbers that are everywhere. These numbers will be the hottest area of concentration in the future for both undergraduate and graduate mathematical research. Both algebraic and figurate aspects of triangular numbers are attractive center of research for mathematicians.

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Open Questions: The following are lists of open questions that we found. 1. Other than 120 is there an other triangular number which can be expressed as the product of five consecutive numbers ? Note : 720: 1*2*3* 4*5 2' . Other 9 (refer to Theorerrl 5) is there any otfrer positive integer n such that 1+ n + n'+ n3+ + nkisatriangularnumber? 3. Other than 6 is there any other positive integer N such that N, 11N, 1 1 1N are all triangular numbers? Note that 6, 66, 666 are triangular numbers.

ACKNOWLEDGEMENT We are indebted to the Editor-in-Chief and the referees for their great assistance. Special thanks to our mentor Dr. Lemma for his guidance by closely working with us. We would like to thank Drs. Jay and Chetty for iheir advice, encouragement, and support. Last but not least we would like to thank PSLSAMP Scholars Program at Savannah State University for providing us with this great opportunity of research expeyierce.

REFEBENCES 1. Jones K, Parker S, and Lemma M: The Mathematical Magic of Perlect Numbers: GaJSci 66(3\ 97 -106, 2008 2. Gupta S: "Fascinating Triangular Numbers" : p3,2002 3. Hamburg C: "Triangular Numbers Are Everywhere!": Illinois, Math- ematics and Science Academy: p5, L992.

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