More about Triangular numbers
Introduction:
The triangular number is a kind of figurate number that can be represented in the form of a triangular grid of dots or unit number 1, where the rows denotes its order n and the total counts up to that row gives the n th triangular number. Simply said , the sum of the natural numbers up to n is the n th triangular number and is denoted by Tn. which can be mathematically expressed as Tn = n (n+1)/2.
Properties of Triangular numbers;
The triangular numbers has many interesting properties. Some of them are
1. All perfect numbers are invariably triangular numbers. First few triangular numbers which are also perfect are 6 (T3) . 28 (T7), 496 (T31), 8128 (T127) 2. The natural series of triangular numbers are continuous alternate blocks of pair of odd and even numbers The sum of two numbers in any block gives even square while one from a block and the neighbouring number in the adjacent block gives odd square. In all cases the sum of any two successive triangular numbers is a square of higher order among them Tn-1 + Tn = n2 and Tn + Tn+1 = (n+1)2. Subtracting one from the other we get an identity (n+1)2 – n2 = Tn+1 - Tn-1 = (2n +1) , which predicts that an odd number can be represented not only as a difference of squares of two numbers with a difference of odd number but also as a difference of two triangular numbers whose orders are one above and below n. The difference between the fourth powers of any two successive numbers gives yet another identity (n+1)4 - n4 = (2n+1) [ Tn+1 + Tn-1 + 2Tn] 3. A square can be expressed in terms of triangular numbers by three ways. n2 = Tn-1 + Tn = T(n2) - T (n2 – 1) = n + 2Tn-1’ It gives an identity T(n2) – Tn = T(n2 -1) – Tn-1
4. With any successive three triangular numbers Tn-1, Tn ,Tn+1 , one can get an identity Tn-1 + T (n2) + Tn+1 = (n+1)2. A kind of discipline is found in such identity with 5,7,9------ successive triangular numbers. Tn-2 + Tn-1 + T (n2) + Tn+1 + Tn+2 = (n+2)2 + 1 Tn-3 + Tn-2 + Tn-1 + T (n2) + Tn+1 + Tn+2 + Tn+3 = (n+3)2 + (1+4) Tn-4 + Tn-3 + Tn-2 + Tn-1 + T (n2) + Tn+1 + Tn+2 +Tn+3 + Tn+4 = (n+4)2 + (1+4+9) In general with m number of triangular numbers above and below the n th triangular number, we get
Tn-m + Tn-m+1 + Tn-m+2 + …………………. Tn2 + Tn+1 + Tn+2 + …………… Tn+m = (n+m)2 + (m-1) ∑ x2 1
5 . Any three triangular numbers with a common difference D in its order shows yet another numeral discipline
D = 1, Tn-1 + (Tn)2 + Tn+1 = (Tn + 1)2 D = 2 , Tn-2 + (Tn)2 + Tn+2 = (Tn + 1)2 + (22 - 1) D = 3 , Tn-3 + (Tn)2 + Tn+3 = (Tn + 1)2 + (32 – 1) In general, when any common difference D = m, Tn-m + (Tn)2 + Tn+m = (Tn + 1)2 + (m2 – 1) 6 Not only square but also any Power of any number cab be expressed in terms of triangular numbers. n3 = (Tn-1 + Tn) n = n + 6 [ n-1 ∑ Tx ] 1 n4 = n2 + 4 Tn Tn-1 n5 = n3 + 6 n2 n-1 ∑ Tx = n3 + 4n Tn-1 Tn
1 n6 = n2 + 12 [ n2 + 3 n-1 ∑ Tx ] [n-1 ∑ Tx] 1 1 7 The Pythagorean triples (a,b,c) satisfying the relation a2 + b2 = c2 have some correlation with its corresponding triangular numbers. Substituting the values for a.b.c in terms of triangular numbers using the property (3) we have . T (c2) – T(b2) – T(a2) = T(c2 -1) – T(b2 – 1) - T (a2 – 1) = (ab)2 Tc – Tb - Ta = Tc-1 – Tb-1 – Ta-1 a + b – c = 2 [ Tc-1 – Tb-1 – Ta-1] It gives an additional condition required for Fermat’s assertion that the relation an + bn = cn where a,b,c are all positive integers and the exponent n > 2 has no whole integral solutions.
T(cn) - T(bn) - T(an) = T(cn- 1) - T(bn-1) – T(an- 1)
8. Innumerable identities can be obtained for every Beal’s expression. Considering the simplest irreducible form of a Beal’s relation 1+ 8 = 9 and using the property (3) we have T1 + T64 – 2T7 – T63 = T81 – 2T8 - T80 9. There are infinitely many square triangular numbers, the first few positive numbers are 1(T1),36(T8) ,1225 (T49),41616 (T288) ,1413721(T1681) ,48024900 (T9800) ,1631432881( T57121),55420693056 (T332928)………… The order and the corresponding square triangular numbers can be determined very
simply from the knowledge of the two product factors of any known square triangular number. Let a2 and b2 are the two product factors of a square triangular number Tx . If a and b are both odd and b>a, then the order of the square triangular number is b2, if a is even and b is odd, then its order will be b2 – 1. One of the two product factors of the next higher order square triangular number would invariably be (a+b)2 and the other product factor depends upon the odd-evenness of the given factors a,b. If both are odd the other product factors is (a+b+ smaller odd number in a,b)2, if one is odd and other is even, the other product factors is (a+b + even number in a,b)2. Conclusion: The identities with triangular numbers can be used as a tool to study relations in number theory
Introduction:
The triangular number is a kind of figurate number that can be represented in the form of a triangular grid of dots or unit number 1, where the rows denotes its order n and the total counts up to that row gives the n th triangular number. Simply said , the sum of the natural numbers up to n is the n th triangular number and is denoted by Tn. which can be mathematically expressed as Tn = n (n+1)/2.
Properties of Triangular numbers;
The triangular numbers has many interesting properties. Some of them are
1. All perfect numbers are invariably triangular numbers. First few triangular numbers which are also perfect are 6 (T3) . 28 (T7), 496 (T31), 8128 (T127) 2. The natural series of triangular numbers are continuous alternate blocks of pair of odd and even numbers The sum of two numbers in any block gives even square while one from a block and the neighbouring number in the adjacent block gives odd square. In all cases the sum of any two successive triangular numbers is a square of higher order among them Tn-1 + Tn = n2 and Tn + Tn+1 = (n+1)2. Subtracting one from the other we get an identity (n+1)2 – n2 = Tn+1 - Tn-1 = (2n +1) , which predicts that an odd number can be represented not only as a difference of squares of two numbers with a difference of odd number but also as a difference of two triangular numbers whose orders are one above and below n. The difference between the fourth powers of any two successive numbers gives yet another identity (n+1)4 - n4 = (2n+1) [ Tn+1 + Tn-1 + 2Tn] 3. A square can be expressed in terms of triangular numbers by three ways. n2 = Tn-1 + Tn = T(n2) - T (n2 – 1) = n + 2Tn-1’ It gives an identity T(n2) – Tn = T(n2 -1) – Tn-1
4. With any successive three triangular numbers Tn-1, Tn ,Tn+1 , one can get an identity Tn-1 + T (n2) + Tn+1 = (n+1)2. A kind of discipline is found in such identity with 5,7,9------ successive triangular numbers. Tn-2 + Tn-1 + T (n2) + Tn+1 + Tn+2 = (n+2)2 + 1 Tn-3 + Tn-2 + Tn-1 + T (n2) + Tn+1 + Tn+2 + Tn+3 = (n+3)2 + (1+4) Tn-4 + Tn-3 + Tn-2 + Tn-1 + T (n2) + Tn+1 + Tn+2 +Tn+3 + Tn+4 = (n+4)2 + (1+4+9) In general with m number of triangular numbers above and below the n th triangular number, we get
Tn-m + Tn-m+1 + Tn-m+2 + …………………. Tn2 + Tn+1 + Tn+2 + …………… Tn+m = (n+m)2 + (m-1) ∑ x2 1
5 . Any three triangular numbers with a common difference D in its order shows yet another numeral discipline
D = 1, Tn-1 + (Tn)2 + Tn+1 = (Tn + 1)2 D = 2 , Tn-2 + (Tn)2 + Tn+2 = (Tn + 1)2 + (22 - 1) D = 3 , Tn-3 + (Tn)2 + Tn+3 = (Tn + 1)2 + (32 – 1) In general, when any common difference D = m, Tn-m + (Tn)2 + Tn+m = (Tn + 1)2 + (m2 – 1) 6 Not only square but also any Power of any number cab be expressed in terms of triangular numbers. n3 = (Tn-1 + Tn) n = n + 6 [ n-1 ∑ Tx ] 1 n4 = n2 + 4 Tn Tn-1 n5 = n3 + 6 n2 n-1 ∑ Tx = n3 + 4n Tn-1 Tn
1 n6 = n2 + 12 [ n2 + 3 n-1 ∑ Tx ] [n-1 ∑ Tx] 1 1 7 The Pythagorean triples (a,b,c) satisfying the relation a2 + b2 = c2 have some correlation with its corresponding triangular numbers. Substituting the values for a.b.c in terms of triangular numbers using the property (3) we have . T (c2) – T(b2) – T(a2) = T(c2 -1) – T(b2 – 1) - T (a2 – 1) = (ab)2 Tc – Tb - Ta = Tc-1 – Tb-1 – Ta-1 a + b – c = 2 [ Tc-1 – Tb-1 – Ta-1] It gives an additional condition required for Fermat’s assertion that the relation an + bn = cn where a,b,c are all positive integers and the exponent n > 2 has no whole integral solutions.
T(cn) - T(bn) - T(an) = T(cn- 1) - T(bn-1) – T(an- 1)
8. Innumerable identities can be obtained for every Beal’s expression. Considering the simplest irreducible form of a Beal’s relation 1+ 8 = 9 and using the property (3) we have T1 + T64 – 2T7 – T63 = T81 – 2T8 - T80 9. There are infinitely many square triangular numbers, the first few positive numbers are 1(T1),36(T8) ,1225 (T49),41616 (T288) ,1413721(T1681) ,48024900 (T9800) ,1631432881( T57121),55420693056 (T332928)………… The order and the corresponding square triangular numbers can be determined very
simply from the knowledge of the two product factors of any known square triangular number. Let a2 and b2 are the two product factors of a square triangular number Tx . If a and b are both odd and b>a, then the order of the square triangular number is b2, if a is even and b is odd, then its order will be b2 – 1. One of the two product factors of the next higher order square triangular number would invariably be (a+b)2 and the other product factor depends upon the odd-evenness of the given factors a,b. If both are odd the other product factors is (a+b+ smaller odd number in a,b)2, if one is odd and other is even, the other product factors is (a+b + even number in a,b)2. Conclusion: The identities with triangular numbers can be used as a tool to study relations in number theory
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