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The Mathematical Beauty of Triangular Numbers

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The Mathematical Beauty of Triangular Numbers 2015 HAWAII UNIVERSITY INTERNATIONAL CONFERENCES S.T.E.A.M. & EDUCATION JUNE 13 - 15, 2015 ALA MOANA HOTEL, HONOLULU, HAWAII S.T.E.A.M & EDUCATION PUBLICATION: ISSN 2333-4916 (CD-ROM) ISSN 2333-4908 (ONLINE) THE MATHEMATICAL BEAUTY OF TRIANGULAR NUMBERS MULATU, LEMMA & ET AL SAVANNAH STATE UNIVERSITY, GEORGIA DEPARTMENT OF MATHEMATICS The Mathematical Beauty Of Triangular Numbers Mulatu Lemma, Jonathan Lambright and Brittany Epps Savannah State University Savannah, GA 31404 USA Hawaii University International Conference Abstract: The triangular numbers are formed by partial sum of the series 1+2+3+4+5+6+7….+n [2]. In other words, triangular numbers are those counting numbers that can be written as Tn = 1+2+3+…+ n. So, T1= 1 T2= 1+2=3 T3= 1+2+3=6 T4= 1+2+3+4=10 T5= 1+2+3+4+5=15 T6= 1+2+3+4+5+6= 21 T7= 1+2+3+4+5+6+7= 28 T8= 1+2+3+4+5+6+7+8= 36 T9=1+2+3+4+5+6+7+8+9=45 T10 =1+2+3+4+5+6+7+8+9+10=55 In this paper we investigate some 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. Most of the results are supplemented with examples. Key Words: Triangular numbers , Perfect square, Pascal Triangles, and perfect numbers. 1. Introduction : The sequence 1, 3, 6, 10, 15, …, n(n + 1)/2, … shows up in many places of mathematics[1] . The Greek called them triangular numbers [1]. The triangular number is a figurate number that can be represented in the form of a triangular grid of points where the first row contains a single element and each subsequent row contains one more element than the previous one as shown below [2]. * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * The Fascinating Mathematical Beauty Of Triangular Numbers * * * * * * * * * * * * * * * * * * * * * 1 3 6 10 15 21 Mathematicians have been fascinated for many years by the properties and patterns of triangular numbers [2]. We can easily hunt for triangular numbers using the formula: nn( 1) Tnn , 0. 2 T1 T2 T3 T4 T5 T6 T7 T8 T9 T10 T11 T12 T13 T14 T15 T16 T17 T18 T19 T20 1 3 6 10 15 21 28 36 45 55 66 78 91 105 120 136 153 171 190 210 The first 20 triangular numbers are as follows. 2. The Main Results: Theorem 1 : Every triangular number is a binomial coefficient . Thoerem2. Every T triangular number is an arithmetic progression. Thoerem 3. Theorem 4 Theorem 5. Every perfect number is a triangular number. Proof without words Refer to the following Pascal’s Triangle [2] and see the red colored numbers. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 2 Lemma , Lambright and Epps nn( 1) n+1 Remark1: Also note that T which is a binomial coefficient for each n 2 2 n 1. Theorem 1: T is a triangular number 8T+1 is a perfect square. Proof: (i) () Assume T is a triangular number. n(n 1) Let T= , n a positive integer. 2 n(n 1) 8T 8 2 n(n 1) 8T 1 8 +1 2 n(n 1) 2 8T 1 8 + 2 2 8nn2 8 2 81T 2 (4n2 4n 1) 8T 1 2 2 = (2n+1)(2n+1) = (2n+1) 2 Hence, 8T+1 is a perfect square. (ii) ( ) Assume 8T+1 is a perfect square. Then 8T+1 is odd for some positive integer n, we have 8T+1= (2n+1) = 4nn2 4 1 T== . Hence, T is a triangular number. By (i) and (ii) the theorem is proved. 3 The Fascinating Mathematical Beauty Of Triangular Numbers Example 1. 6 is a triangular number 8(6)+1 =49 is a perfect square Example 2. 8(15)+1= 121, a perfect square 15 is a triangular number. 8T 1 1 Corollary 1. T is a triangular number n = is an integer. 2 Proof: The corollary easily follows by Theorem 1. Theorem 2: If Tm and Tn are triangular numbers, then Tm n T m T n mn for m and n positive integers. Proof: m(m 1) nn( 1) Note: Tm & Tn . Then 2 2 m( m 1) n ( n 1) T T mn +mn m n 22 m2 m n2 n = mn 2 m2 m n2 n 2mn m2 2mn n2 m n = = 2 2 (m n)(m n) (m n) (m n)m n 1 = = = T 2 2 mn Example 3. Consider T34 and T . Note that T34= 6 and T 10 . Observe that T34 T7 =28 and T34 + T 3(4) 6 10 12 28 . Hence, T34 = T34 + T +3(4) Theorem 3: If Tm and Tn are triangular numbers, then TTTTTmn m n m11 n m(m 1) n(n 1) Proof: Note: T and T . Then m 2 n 2 4 Lemma , Lambright and Epps m(m 1) n(n 1) (m 1)m (n 1)n T T T T = m n m1 n1 2 2 2 2 m 2 m n 2 n m 2 m n 2 n = 2 2 2 2 m 2 n 2 mn 2 nm2 mn m 2 n 2 mn 2 nm2 mn = + n(n 1) 4 4 2 2m2n2 2mn 2mn (mn 1) mn (mn 1) = = = 4 4 2 =T mn Example 4. Let m =6 and n =7. Then T6721 and T 28. nn( 1) 42(43) By using Tn , we get T(6)(7) =T 42 903 22 We also have TTTTT5 615(21) 315 and T 6 7 5 6 588+315=903. Hence,T(6)(7) = TTTTT42 6 7 5 6 Lemma 1. The sum of two consecutive triangular numbers is a perfect square Proof: Let T n1 and Tn be any two consecutive triangular numbers, such that (n 1)(n) Tn1 = and = 2 Then, (n 1)n n(n 1) TT = nn1 2 2 5 The Fascinating Mathematical Beauty Of Triangular Numbers n2 n n2 n 2n 2 = = = n2 2 2 which is a perfect square. Example 5. Let T6 and T 7 be any consecutive triangular numbers. Then TT67 + 21+28= 49, which is a perfect square. k( k 1)(2 k 1) Lemma 2. 12 2 2 3 2 4 2 ... k 2 6 Proof. We can easily prove the lemma using induction. Example 6. Let k = 5 . Then 12 2 2 3 3 4 2 5 2 149162555 . 5(6)(11) 5(6)(11) Also, we have 55 and hence 12 2 2 3 3 4 2 5 2 . 6 6 Theorem 4. If Tk be triangular numbers fork k 0 , then we have n n( n 1)( n 2) Tk k1 6 Proof: To prove the theorem, we apply divide and conquer method by considering two cases: (1) If n is even, say n =2k, then TTTTTTTTT ... ( ) ( ) ... ( ) 1 2n 1 2 3 4 2 k 1 2 k = 22 4 2 ... (2k ) 2 ( by Lemma 1) = 4(12 2 2 ... k 2 ) 4k (2 k 1)( k 1) = (by Lemma 2). 6 n( n 1)( n 2) = as n = 2k 6 6 Lemma , Lambright and Epps (2) If n is odd, say n= 2k+1, then TTTTTTTTTT... ( ) ( ) ... ( ) 12n 1234 21221 k k k (2kk 1)(2 2) = + ( by Lemma 1 and definition of ) 2 = + 3(2kk 1)(2 2) = + ( by Lemma 2.) 6 2k (2 k 1)(2 k 2) = + 6 (2k 1)(2 k 2)(2 k 3) = 6 Tk = as n = 2k+1 By (1) and (2) the Theorem is proved. 5 Example 7. Let n = 5. Then TTTTTTk 1 2 3 4 5 = 1+3+6+10+15= 35. We also have, k1 5(6)(7) 5 5(6)(7) 35 and hence Tk . 6 k1 6 Theorem 5 For any natural number n, the number 22 4 2 ... (2k ) 2 1 + 9 + 92 + 93 + ... + 9n is a triangular number. 4(12 2 2 ... k 2 ) 2 3 n 91n1 Proof: Let T =1 + 9 + 9 + 9 + ... + 9 . Then T= . 4k (2 k 1)( k 1) 8 By Theorem 1 it is suffice to prove that is a perfect square. We will apply the divide , 6 8T+1 and conquer method as in Theorem 4. n( n 1)( n 2) 6 (1) If n is even, say n =2k, then 7 The Fascinating Mathematical Beauty Of Triangular Numbers n1 2 91 n1 21k 2kk 2 1 8T+1 = 8 ( )1 = 9 =9 = 9 9 3 , which a perfect square. 8 (2) If n is odd, say n= 2k+1, then 2 8T+1 = 8 ( = = 922k = 9k 1 , which a perfect square. By (1)and (2), the theorem is proved. Tn Theorem 6. If be triangular numbers for n 1, then we have 1 = 2 n1 Tn Proof: 2 n1 nn( 1) 11 2 n1 nn1 = 2(1) =2 Proposition 2. The difference of the squares of two consecutive triangular numbers is a cube. (n 1)n (n 1)n Proof: Consider = and Tn = 2 2 22 22n( n 1 ( n 1) n Then, ()()TTnn 1 = 22 T n 1 n4 2n2 n2 n4 2n3 n2 4n3 3 = - = = n 4 4 4 Example 8. Let T67 and T be any two consecutive triangular numbers.
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