Download triangular numbers—some general theorems and related results

TRIANGULAR NUMBERS—SOME GENERAL
THEOREMS AND RELATED RESULTS
R. Narayanan
Dedicated to my esteemed teacher, the inimitable N. B. N. – SSKZM
Abstract. The following results are proved.
If T = {T0 , T1 , T2 , T3 , - - -Tn , - - -, ∞}
[T0 = 0] are Triangu-
lar numbers
1. then (2ξ + 1)2 Tn + Tξ
[ξ ≥ 0, n ≥ 0] are also Triangular
Numbers.
2. then (2λ)2 Tn + Tλ + λn[λ ≥ 1, n ≥ 0] are also Triangular
numbers.
3. The square of every odd number can be expressed as the
difference of two Triangular Numbers.
4. The square of every λth even number can be expressed as the
difference of the ‘(3λ)th Triangular Number’ and ‘the sum of
the λth Triangular Number and λ’.
2
5. For any Natural Number N ≥ 2, there Exist Infinite Triangular Numbers that are the Sum of N Triangular Numbers
each.
6. If
n(n + 2α + 1)
n(n + 2α + 1)
+ 1, y = n + 1, z =
2
2
Tx = Ty + Tz + (α − 1)n = Ty + Tz + (α − 1)(y − 1)
x=
[α = 0, 1, 2, - - -∞] [n = 0, 1, 2, - - -, ∞]
7. If
(n + 1)(n + 2β)
(n + 1)(n + 2β)
+ 1, y = n + 1, z =
2
2
Tx = Ty + Tz + β + (β − 1)n = Ty + Tz + β
x=
+ (β − 1)(y − 1)[β = 1, 2, 3 - - -∞] [n = 0, 1, 2, - - -, ∞]
Various related sub-results are also proved.
Infinite Types of unique sub-classes of Triangular Numbers
are defined and the formulae or their sum upto “n” terms are
derived and these are expressed in terms of binomial coefficients
also.
Key words:- Triangular Numbers, Binomial Coefficiets, Infinite
Generalizations.
AMS Subject Classification No:- 11A99.
Triangular Numbers—Some General Theorems · · ·
1
3
Some General Results
Theorem 1.1. If T = {T0 , T1 , T2 , T3 , - - -Tn , - - -∞}[T0 = 0] are Triangular Numbers then (2ξ + 1)2 Tn + Tξ [ξ ≥ 0, n ≥ 0] are also Triangular
Numbers.
[Euler’s results that (9Tn + 1), (25Tn + 3), (49Tn + 6), [n ≥ 0] are
Triangular, are obtained as special cases when we set ξ = 1, 2, and 3
respectively in the formula (2ξ + 1)2 Tn + Tξ .]
Proof. We prove the result by repeated induction.
For ξ = 0, (2ξ + 1)2 Tn + Tξ are the set of Triangular Numbers themselves. For ξ = 1, 2, 3 we have the standard results due to Euler which
are proved by induction on “n” for each specific value of ξ. To prove the
general formula for all values of ξ ≥ 0 we assume that the result is true
for ξ = k and prove that it is true for ξ = k + 1 and thus close the doubly
(infinitely) inductive argument. We state the complete set of Triangular
Numbers for each value of ξ as follows which can easily be proved by
induction and the Pythagorean. Theorem that a number is Triangular if
n(n + 1)
and only if it is of the form
, in each case.
2
[Note on the notation:
[R]
L
In Tn , R may be called the Root Level and L the Layer level which
define a specific sub-class of Triangular Numbers.]
For ξ = 0, 2ξ + 1 = 1
[1]
n(n + 1)
= Tn + T0 = (T0 + T1 )Tn + T0 = Tn
Tn =
2
1
[n ≥ 1]
4
For ξ = 1, 2ξ + 1 = 3
[3]
(3n + 1)(3n + 2)
Tn =
= 9Tn + T1 = (T2 + T3 )Tn + T1
2
= T(3n+1) [n ≥ 1]
1
For ξ = 2, 2ξ + 1 = 5
[3]
(6n + 1)(6n + 2)
Tn =
= 25Tn + T2 = (T4 + T5 )Tn + T2
2
= T(6n+1) [n ≥ 1]
2
For ξ = 3, 2ξ + 1 = 7
[3]
(9n + 1)(9n + 2)
= 49Tn + T3 = (T6 + T7 )Tn + T3
Tn =
2
= T(9n+1) [n ≥ 1]
3
For ξ = 4, 2ξ + 1 = 9
[3]
(12n + 1)(12n + 2)
Tn =
= 81Tn + T4
2
= (T8 + T9 )Tn + T4 = T(12n+1) [n ≥ 1]
4
For ξ = 5, 2ξ + 1 = 11
[3]
(15n + 1)(15n + 2)
= 121Tn + T5
Tn =
2
= (T10 + T11 )Tn + T5 = T(15n+1) [n ≥ 1]
5
Triangular Numbers—Some General Theorems · · ·
5
For ξ = 6, 2ξ + 1 = 13
[3]
(18n + 1)(18n + 2)
Tn =
= 169Tn + T6
2
= (T12 + T13 )Tn + T6 = T(18n+1)
[n ≥ 1]
6
Clearly if ξ = k
[3]
(3kn + 1)(3kn + 2)
k
Tn =
= (2k + 1)2 Tn + Tk
2
= (T2k + T2k+1 ) + Tn + Tk = T(3kn+1) [n ≥ 1])
are Triangular, then for ξ = k + 1
[3]
(3(k + 1)n + 1)(3(k + 1)n + 2)
k + 1Tn =
2
= (2(k + 1) + 1)2 Tn + Tk+1
= (T2(k+1) + T2(k+1)+1 ) + Tn + Tk+1
= T(3(k+1)n+1)
[n ≥ 1]
are also Triangular thus proving that
ξ
(3ξn + 1)(3ξn + 2)
[3]
=
= (2ξ + 1)2 Tn + Tξ
Tn
2
= (T2ξ + T2ξ+1 )Tn + Tξ = T(3ξn+1)
[ξ ≥ 1, n ≥ 1]
(1.1)
are all Triangular. Thus we may say that Triangular Numbers derived
6
form Root Level 3 Layer Levels ξ[ξ ≥ 1] are Eulerian classes of Triangular Numbers and we may call “ξ” the Eulerian Class Level Number.
For ξ = 0 the Eulerian class is the set of all Triangular Numbers.
Corollary 1.1. Every Triangular Number can be Expressed the Difference of a Triangular Number and the Sum of two Triangular numbers
and the product of another Triangular Number in Infinite Ways.
Tξ = T(3ξn+1) − (T2ξ + T2ξ+1 )Tn
[ξ ≥ 1, n ≥ 1]
Proof. Trivial from (1.1).
Corollary 1.2. Every Triangular Number can be Expressed as the Ratio
of the Difference of Two Triangular Numbers and the Sum of Two Triangular Numbers in Infinite Ways.
T(3ξn+1) − Tξ
Tn =
(T2ξ + T2ξ+1 )
[ξ ≥ 1, n ≥ 1]
Proof. Trivial form (1.1).
Corollary 1.3. For ξ ≥ 1, n ≥ 1
(2ξ + 1)2 (n + 1)2 = T(3ξn+1) + T(3ξ(n+1)+1) − 2Tξ
Proof.
[3]
(3ξn + 1)(3ξn + 2)
ξ Tn =
= (2ξ + 1)2 Tn + Tξ
2
= (T2ξ + T2ξ+1 )Tn + Tξ = T(3ξn+1) [ξ ≥ 1, n ≥ 1]
(1.2)
Triangular Numbers—Some General Theorems · · ·
(2ξ + 1)2 Tn + Tξ = T(3ξn+1) .
7
(1.3)
Substituting n = n + 1 in (1.3)
(2ξ + 1)2 Tn+1 + Tξ = T(3ξ(n+1)+1)
(1.4)
adding the two equations, substituting Tn + Tn+1 = (n + 1)2 [Nicomachus] and simplifying we have
(2ξ + 1)2 (n + 1)2 = T(3ξn+1) + T(3ξ(n+1)+1) − 2Tξ
[ξ ≥ 1, n ≥ 1]
(1.5)
2ξ + 1 defines the set of all odd numbers. Seeking a symmetric result
for all even numbers, we have the following theorem.
Theorem 1.2. If T = {T0 , T1 , T2 , T3 , - - -Tn , - - -∞}[T0 = 0] are Triangular Numbers then (2λ)2 Tn + Tλ + λn [λ ≥ 1, n ≥ 0] are also
Triangular Numbers.
Proof. We prove the result by repeated induction.
First we prove for λ = 1, 2, 3 by induction on “n” for each specific value
of λ. To prove the general formula for all values of λ ≥ 1 we assume
that the result is true for λ = k and prove that it is true for λ = k + 1
and thus close the doubly (infinitely) inductive argument. We state the
complete set of Triangular Numbers for each value of λ as follows which
can easily be proved by induction and the Pythagorean Theorem that a
n(n + 1)
, in each
number is Triangular if and only if if it is the form
2
case.
8
For λ = 1, 2λ = 2
[1]
2 T = (2n + 1)(2n + 2) = 4T + T + 1n
n
n
1
2
= (T1 + T2 )Tn + T1 + 1(n) = T3n [n ≥ 1]
For λ = 2, 2λ = 4
[1]
5 T = (5n + 1)(5n + 2) = 16T + T + 2n
n
n
2
2
= (T3 + T4 )Tn + T2 + 2n = T6n [n ≥ 1]
For λ = 3, 2λ = 6
[1]
8 T = (8n + 1)(8n + 2) = 36T + T + 3n
n
n
3
2
= (T5 + T6 )Tn + T3 + 3n = T9n [n ≥ 1]
For λ = 4, 2λ = 8
[1]
11 T = (11n + 1)(11n + 2) = 64T + T + 4n
n
n
4
2
= (T7 + T8 )Tn + T4 + 4n = T12n [n ≥ 1]
For λ = 5, 2λ = 10
[1]
14 T = (14n + 1)(14n + 2) = 100T + T + 5n
n
5
n
2
= (T9 + T10 )Tn + T5 + 5n = T15n
Triangular Numbers—Some General Theorems · · ·
9
For λ = 6, 2λ = 12
[1]
17 T = (17n + 1)(17n + 2) = 144T + T + 6n
n
n
6
2
= (T11 + T12 )Tn + T6 + 6n = T18n [n ≥ 1]
For λ = 7, 2λ = 14
[1]
20 T = (20n + 1)(20n + 2) = 196T + T + 7n
n
n
7
2
= (T13 + T14 )Tn + T7 + 7n = T21n [n ≥ 1]
For λ = k, 2λ = 2k
[1]
(3k − 1) T = ((3k − 1)n + 1)((3k − 1)n + 2)
n
2
2
= (2k) Tn + Tk + kn
= (Tk−1 + Tk )Tn + Tk + kn = T3kn
[n ≥ 1]
For λ = k + 1, 2λ = 2(k + 1)
[1]
(3k + 2) T = ((3k + 2)n + 1)((3k + 2)n + 2)
n
2
= (2(k + 1))2 Tn + T(k+1) + (k + 1)n
= (Tk + Tk+1 )Tn + T(k+1) + (k + 1)n = T(3k+1)n
[n ≥ 1]
10
are also triangular thus proving that
[1]
(3λ − 1) T = ((3λ − 1)n + 1)((3λ − 1)n + 2)
n
2
2
= (2λ) Tn + Tλ + λn
= (T2λ−1 + T2λ )Tn + Tλ + λn = T3λn
(1.6)
[n ≥ 1]
are all Triangular. Thus we may say that Triangular Numbers derived
from Root Level 1 Layer Levels (3λ−1)[λ ≥ 1] are Pythagorean classes
of Triangular Numbers and we may call “(3λ − 1)” the Pythagorean
Class Level Number.
Corollary 1.4. Every Triangular Number can be Expressed the “Difference of a Triangular Number” and “the Sum of Two Triangular Numbers and the product of another Triangular Number and a specific constant (product of the two indexing factors)” in Infinite Ways.
Tλ = T3λn − (T2λ−1 + T2λ )Tn − λn
[λ ≥ 1, n ≥ 1]
Proof. Trivial from (1.6)
Corollary 1.5. Every Triangular Number can be Expressed as the Ratio
of the “Difference of a Triangular Number and the Sum of a Triangular
Number and a specific constant (product of the two indexing factors)”
and the “Sum of Two Triangular Numbers” in Infinite Ways.
Tλ = T3λn − (T2λ−1 + T2λ )Tn − λn
Tn =
T3λn − (Tλ + λn)
(T2λ−1 + T2λ )
[λ ≥ 1, n ≥ 1]
Triangular Numbers—Some General Theorems · · ·
11
Proof. Trivial from (1.6)
Corollary 1.6. For λ ≥ 1, n ≥ 1
(2λ)2 (n + 1)2 = T3λn + T(3λ(n+1)) − (2Tλ + λ(2n + 1)).
(1.7)
Proof.
[1]
(3λ − 1) T = ((3λ − 1)n + 1)((3λ − 1)n + 2)
n
2
= (2λ)2 Tn + Tλ + λn
(1.8)
= (T2λ−1 + T2λ )Tn + Tλ + λn = T3λn
[λ ≥ 1, n ≥ 1]
(2λ)2 Tn + Tλ + λn = T3λn
(1.9)
Substituting n = n + 1 in (1.9)
(2λ)2 Tn+1 + Tλ + λ(n + 1) = T(3λ(n+1))
(1.10)
adding the two equations substituting Tn + Tn+1 = (n + 1)2 [Nicomachus] and simplifying we have
(2λ)2 (n + 1)2 = T3λn + T(3λ(n+1)) − (2Tλ + λ(2n + 1))
[λ ≥ 1, n ≥ 1] (1.11)
12
2
Two General Theorems on Triangular Numbers
Theorem 2.1. The square of every odd number can be expressed as the
difference of two Triangular Numbers.
Proof. We have
(2ξ + 1)2 Tn + Tξ = T(3ξn+1)
(2.1)
Substituting n = 1 in (2.1), we have
(2ξ + 1)2 = T(3ξ+1) − Tξ
[ξ ≥ 0]
(2.2)
The standard result that “the square of any odd multiple of 3 is the difference of two Triangular Numbers [i.e., 9(2n + 1)2 = T9n+4 − T3n+1 ]”
[Ref. 6], clearly is a cumbersome expression of a special case when the
“odd multiples of 3” are the odd numbers concerned! Again, looking for
General Symmetry we have.
Theorem 2.2. The square of every λth even number can be expressed as
the difference of the ‘(3λ)th , Triangular Number’ and ‘the sum of the λth
Triangular Number and λ’.
Proof. We have,
(2λ)2 Tn + Tλ + λn = T3λn
(2.3)
Substituting n = 1 in (2.3), we have
(2λ)2 = T3λ − (Tλ + λ)
[λ ≥ 1]
(2.4)
Triangular Numbers—Some General Theorems · · ·
3
13
Infinite Root Levels and Layer Levels of Triangular Numbers
In this section we define various Root-Levels and Layer-Levels of Triangular Numbers and investigate them.
[R]
In L Tn , R may be called the Root Level and L the Layer Level which
define a specific sub-class of Triangular Numbers.
The sequence of Layer-Levels for each Root-Level Facilitates the
Proof for All Root-Levels.
[1]
(n + 1)(n + 2)
1T
n+1 =
2
[n = 0, 1, 2, - - -, ∞]
= T1 , T2 , T3 , T4 , T5 , - - -, T1+(n−1) , - - -, ∞
[1]
(2n + 1)(2n + 2)
2T
n+1 =
2
[n = 0, 1, 2, - - -, ∞]
= T1 , T3 , T5 , T7 , T9 , - - -, T1+2(n−1) , - - -, ∞
[1]
(3n + 1)(3n + 2)
3T
n+1 =
2
[n = 0, 1, 2, - - -, ∞]
= T1 , T4 , T7 , T10 , T13 , - - -, T1+3(n−1) , - - -, ∞
[1]
(4n + 1)(4n + 2)
4T
n+1 =
2
[n = 0, 1, 2, - - -, ∞]
= T1 , T5 , T9 , T13 , T17 , - - -, T1+4(n−1) , - - -, ∞
−−−−−−−−−
14
[1]
(φn + 1)(φn + 2)
φT
n+1 =
2
[n = 0, 1, 2, - - -, ∞]
= T1 , T1+φ , T1+2φ , T1+3φ , T1+4φ , - - -, T1+φ(n−1) , - - -, ∞
[2]
(2n + 1)(2n + 2)
1T
n+1 =
2
[n = 0, 1, 2, - - -, ∞]
= T1 , T3 , T5 , T7 , T9 , - - -, T1+2(n−1) , - - -, ∞
[2]
(4n + 1)(4n + 2)
2T
n+1 =
2
[n = 0, 1, 2, - - -, ∞]
= T1 , T5 , T9 , T13 , T17 , - - -, T1+4(n−1) , - - -, ∞
[2]
(6n + 1)(6n + 2)
3T
n+1 =
2
[n = 0, 1, 2, - - -, ∞]
= T1 , T7 , T13 , T19 , T25 , - - -, T1+6(n−1) , - - -, ∞
[2]
(8n + 1)(8n + 2)
4T
n+1 =
2
[n = 0, 1, 2, - - -, ∞]
= T1 , T9 , T17 , T25 , T33 , - - -, T1+8(n−1) , - - -, ∞
−−−−−−−−−−−
[2]
(2φn + 1)(2φn + 2)
φT
n+1 =
2
[n = 0, 1, 2, - - -, ∞]
= T1 , T1+2φ , T1+4φ , T1+6φ , T1+8φ , - - -, T1+(2φn−1) , - - -, ∞
Triangular Numbers—Some General Theorems · · ·
[3]
(3n + 1)(3n + 2)
1T
n+1 =
2
15
[n = 0, 1, 2, - - -, ∞]
= T1 , T4 , T7 , T10 , T13 , - - -, T1+3(n−1) , - - -, ∞
[3]
(6n + 1)(6n + 2)
2T
n+1 =
2
[n = 0, 1, 2, - - -, ∞]
= T1 , T7 , T13 , T19 , T25 , - - -, T1+6(n−1) , - - -, ∞
[3]
(9n + 1)(9n + 2)
3T
n+1 =
2
[n = 0, 1, 2, - - -, ∞]
= T1 , T10 , T19 , T28 , T37 , - - -, T1+9(n−1) , - - -, ∞
[3]
(12n + 1)(12n + 2)
4T
n+1 =
2
[n = 0, 1, 2, - - -, ∞]
= T1 , T13 , T25 , T37 , T49 , - - -, T1+12(n−1) , - - -, ∞
− − − − − − −−
[3]
(3φn + 1)(3φn + 2)
φT
n+1 =
2
[n = 0, 1, 2, - - -, ∞]
= T1 , T1+3φ , T1+6φ , T1+9φ , T1+12φ , - - -, T1+3φ(n−1) , - - -, ∞
[4]
(4n + 1)(4n + 2)
1T
n+1 =
2
[n = 0, 1, 2, - - -, ∞]
= T1 , T5 , T9 , T13 , T17 , - - -, T1+4(n−1) , - - -, ∞
16
[4]
(8n + 1)(8n + 2)
2T
n+1 =
2
[n = 0, 1, 2, - - -, ∞]
= T1 , T9 , T17 , T25 , T33 , - - -, T1+8(n−1) , - - -, ∞
[4]
(12n + 1)(12n + 2)
3T
n+1 =
2
[n = 0, 1, 2, - - -, ∞]
= T1 , T13 , T25 , T37 , T49 , - - -, T1+12(n−1) , - - -, ∞
[4]
(16n + 1)(16n + 2)
4T
n+1 =
2
[n = 0, 1, 2, - - -, ∞]
= T1 , T17 , T33 , T49 , T65 , - - -, T1+16(n−1) , - - -, ∞
−−−−−−−−−
[4]
(φ4n + 1)(φ4n + 2)
φT
n+1 =
2
[n = 0, 1, 2, - - -, ∞]
= T1 , T1+4φ , T1+8φ , T1+12φ , T1+16φ , - - -, T1+4φ(n−1) , - - -, ∞
− − − − − − −−
[R]
(Rn + 1)(Rn + 2)
1T
n+1 =
2
[n = 0, 1, 2, - - -, ∞]
= T1 , T1+R , T1+2R , T1+3R , T1+4R , - - -, T1+R(n−1) , - - -, ∞
[R]
(2Rn + 1)(2Rn + 2)
2T
n+1 =
2
[n = 0, 1, 2, - - -, ∞]
= T1 , T1+2R , T1+4R , T1+6R , T1+8R , - - -, T1+2R(n−1) , - - -, ∞
Triangular Numbers—Some General Theorems · · ·
[R]
(3Rn + 1)(3Rn + 2)
3T
n+1 =
2
17
[n = 0, 1, 2, - - -, ∞]
= T1 , T1+3R , T1+6R , T1+9R , T1+12R , - - -, T1+3R(n−1) , - - -, ∞
[R]
(4Rn + 1)(4Rn + 2)
4T
n+1 =
2
[n = 0, 1, 2, - - -, ∞]
= T1 , T1+4R , T1+8R , T1+12R , T1+16R , - - -, T1+4R(n−1) , - - -, ∞
− − − − − − −−
[R]
(φRn + 1)(φRn + 2)
φT
n+1 =
2
[n = 0, 1, 2, - - -, ∞]
= T1 , T1+φR , T1+2φR , T1+3φR , T1+4φR , - - -, T1+φR(n−1) , - - -, ∞
[φ = 1, 2, - - -∞]
[R = 1, 2, - - -, ∞]
4
The Infinite Triangular Number Classes in terms of Binomial
Coefficients
We can easily prove by Repeated Induction the following results.
[n = 0, 1, 2, - - -, ∞]
[1]
n
+
2
1T
n+1 =
2
[1]
2n + 2
2T
n+1 =
2
= Tn + 1 [This is the standard result. Ref.6)]
18
[1]
3n + 2
3T
n+1 =
2
[1]
4n
+
2
4T
n+1 =
2
−−−−−−−
[1]
φn
+
2
φT
n+1 =
2
[2]
2n + 2
1T
n+1 =
2
[2]
4n
+
2
2T
n+1 =
2
[2]
6n + 2
3T
n+1 =
2
[2]
8n
+
2
4T
n+1 =
2
− − − − −−
[2]
2φn
+
2
φT
n+1 =
2
[3]
3n + 2
1T
n+1 =
2
Triangular Numbers—Some General Theorems · · ·
[3]
6n + 2
2T
n+1 =
2
[3]
9n
+
2
3T
n+1 =
2
[3]
12n + 2
4T
n+1 =
2
− − − − − − −−
[3]
3φn
+
2
φT
n+1 =
2
[4]
4n
+
2
1T
n+1 =
2
[4]
8n
+
2
2T
n+1 =
2
[4]
12n + 2
3T
n+1 =
2
[4]
16n
+
2
4T
n+1 =
2
−−−−−−−
[4]
4φn + 2
φT
n+1 =
2
19
20
−−−−−−−
The result if True for k ie.
[k]
[k + 1]
kφn
+
2
φT
φ T
⇒
n+1 =
n+1
2
(k + 1)φn + 2
=
2
is also True for k + 1 proving the General Result for R = 1, 2, - - -∞
[R]
Rn
+
2
1T
n+1 =
2
[R]
2Rn
+
2
2T
n+1 =
2
[R]
3Rn
+
2
3T
n+1 =
2
[R]
4Rn
+
2
4T
n+1 =
2
− − − − −−
[R]
φRn
+
2
φT
n+1 =
2
[φ = 1, 2, - - -, ∞]
[R = 1, 2, - - -, ∞]
Triangular Numbers—Some General Theorems · · ·
5
21
The Formulae for the Sum up to the nth term of Infinite Root
Levels and Layer Levels of Triangular Numbers
In this section we derive the Formulae for the Sum up to the nth term of
Infinite Root Levels and Layer Levels of Triangular Numbers. Aryabhatta’s traditional formula for the Sum of Triangular Numbers is assumed.
For this purpose we need to consider a simple case of General Arithmetic cum Arithmetic Inductive Progressions.
Let
[P]n = a + b(n − 1) +
c(n − 1)(n − 2)
2
[Ref.7]
(5.1)
[a, b and c are algebraic numbers.]
[Substituting a = 1, b = 2 and c = 1 we get the sequence of the set of
All Triangular Numbers T.]
Now we can easily prove by Induction that the sum of this progression up to n terms is
n(n − 1)
Sn [P] = an + b
2
c(n − 1)(n − 2) 2c(n − 2)(n − 3)
+
+
2
2
3c(n − 1)(n − 2)
(n − 2)c(2)(1)
+
...
.
2
2
22
n−2
X
n(n − 1)
(n − i)(n − (i + 1))
= an + b
+c
i
2
2
1
n(n − 1)
n(n − 1)(n − 2)
= an + b
+c
2!
3!
b
n
c
n
a n
=
+
+
1
2
1
3
1
1
n≥1
n≥2
n≥3
n−1 [1]
n−1
X
X
(i + 1)(i + 2)
1T
=
i+1
2
(5.2)
[n = 1, 2, - - -, ∞]
0
0
=
n
X
1
n(n − 1)
T1 + (i − 1) = n + (T2 − T1 )
2
n−2
X
(n − i)(n − (i + 1))
2
+1
i
2
1
Substituting a = 1, b = 2, c = 12 = 1, in (5.2) we have
X
n−2
(n − i)(n − (i + 1))
n(n − 1)
=n+2
+
i
2
2
1
n(n − 1)
n(n − 1)(n − 2)
=n+2
+
2!
3!
n
n
2 n
=
+
+
3
1
1
2
n≥1
n≥2
n≥3
Triangular Numbers—Some General Theorems · · ·
n−1 [1]
n−1
X
X
(2i + 1)(2i + 2)
2T
=
i+1
2
23
[n = 1, 2, - - -, ∞]
0
0
=
n
X
T1 + 2(i − 1)
1
n−2
X
n(n − 1)
(n − i)(n − (i + 1))
2
= n + (T3 − T1 )
+2
i
2
2
1
Substituting a = 1, b = 5, c = 22 = 4, in (5.2) we have
n−2
X
n(n − 1)
(n − i)(n − (i + 1))
=n+5
+4
i
2
2
1
n(n − 1)
n(n − 1)(n − 2)
=n+5
+4
2!
3!
n
5 n
4 n
=
+
+
1
1
1
2
3
n≥1
n≥2
n≥3
n−1 [1]
n−1
X
X
(3i + 1)(3i + 2)
3T
i+1 =
2
0
[n = 1, 2, - - -, ∞]
0
=
n
X
1
n(n − 1)
T1 + 3(i − 1) = n + (T4 − T1 )
2
n−2
X
(n − i)(n − (i + 1))
+32
i
2
1
Substituting a = 1, b = 9, c = 32 = 9, in (5.2) we have
24
n−2
X
(n − i)(n − (i + 1))
n(n − 1)
i
+9
=n+9
2
2
1
n(n − 1)
n(n − 1)(n − 2)
=n+9
+9
2!
3!
9 n
4 n
n
+
+
=
1
2
1
3
1
n≥2
n≥3
n≥1
n−1
n−1 [1]
X
X
(4i + 1)(4i + 2)
4T
=
i+1
2
[n = 1, 2, - - -, ∞]
0
0
=
n
X
1
n(n − 1)
T1 + 4(i − 1) = n + (T5 − T1 )
2
n−2
X
(n − i)(n − (i + 1))
2
+4
i
2
1
Substituting a = 1, b = 14, c = 42 = 16, in (5.2) we have
n−2
X
(n − i)(n − (i + 1))
n(n − 1)
+ 16
i
= n + 14
2
2
1
n(n − 1)
n(n − 1)(n − 2)
+ 16
= n + 14
2!
3!
n
14 n
16 n
=
+
+
1
1
2
1
3
n≥1
n≥2
n≥3
−−−−−−−
Triangular Numbers—Some General Theorems · · ·
n−1
X
0
[1]
n−1
X
(φi + 1)(φi + 2)
φT
i+1 =
2
25
[n = 1, 2, - - -, ∞]
0
=
n
X
1
n(n − 1)
T1 + φ(i − 1) = n + (T1 + φ − T1 )
2
n−2
X
(n − i)(n − (i + 1))
2
i
+φ
2
1
(φ + 1)(φ + 2)
− 1 , c = φ2 , in (5.2) we have
Substituting a = 1, b =
2
(φ + 1)(φ + 2)
n(n − 1)
=n
−1
2
2
n−2
X
(n − i)(n − (i + 1))
2
+φ
i
2
1
(φ + 1)(φ + 2)
n(n − 1)
=n+
−1
2
2!
2 n(n − 1)(n − 2)
+φ
3!
(φ + 1)(φ + 2) − 1 2 n
φ
n
n
2
=
+
+
1
2
1
3
1
n≥1
n≥2
n≥3
n−1 [2]
n−1
X
X
(2i + 1)(2i + 2)
1T
=
i+1
2
0
[n = 1, 2, - - -, ∞]
0
=
n
X
1
n(n − 1)
T1 + 2(i − 1) = n + (T3 − T1 )
2
26
n−2
X
(n − i)(n − (i + 1))
i
+2
2
1
2
Substituting a = 1, b = 5, c = 22 = 4, in (5.2) we have
n−2
X
(n − i)(n − (i + 1))
n(n − 1)
i
+4
=n+5
2
2
1
n(n − 1)
n(n − 1)(n − 2)
=n+5
+4
2!
3!
4 n
n
5 n
=
+
+
1
3
1
1
2
n≥1
n≥2
n≥3
n−1 [2]
n−1
X
X
(4i + 1)(4i + 2)
2T
i+1 =
2
0
[n = 1, 2, - - -, ∞]
0
=
n
X
1
n(n − 1)
T1 + 4(i − 1) = n + (T5 − T1 )
2
n−2
X
(n − i)(n − (i + 1))
2
+4
i
2
1
Substituting a = 1, b = 14, c = 42 = 16, in (5.2) we have
n−2
X
n(n − 1)
(n − i)(n − (i + 1))
= n + 14
+ 16
i
2
2
1
n(n − 1)
n(n − 1)(n − 2)
= n + 14
+ 16
2!
3!
Triangular Numbers—Some General Theorems · · ·
27
n
14 n
16 n
=
+
+
1
1
2
1
3
n≥1
n≥2
n≥3
n−1
n−1 [2]
X
X
(6i + 1)(6i + 2)
3T
i+1 =
2
[n = 1, 2, - - -, ∞]
0
0
=
n
X
1
n(n − 1)
T1 + 6(i − 1) = n + (T7 − T1 )
2
n−2
X
(n − i)(n − (i + 1))
2
i
+6
2
1
Substituting a = 1, b = 27, c = 62 = 36, in (5.2) we have
n−2
X
(n − i)(n − (i + 1))
n(n − 1)
= n + 27
+ 36
i
2
2
1
n(n − 1)
n(n − 1)(n − 2)
= n + 27
+ 36
2!
3!
n
27 n
36 n
=
+
+
1
1
1
2
3
n≥1
n≥2
n≥3
n−1 [2]
n−1
X
X
(8i + 1)(8i + 2)
4T
=
i+1
2
0
[n = 1, 2, - - -, ∞]
0
=
n
X
1
n(n − 1)
T1 + 8(i − 1) = n + (T9 − T1 )
2
n−2
X
(n − i)(n − (i + 1))
2
+8
i
2
1
28
Substituting a = 1, b = 44, c = 82 = 64, in (5.2) we have
n−2
X
(n − i)(n − (i + 1))
n(n − 1)
i
+ 64
= n + 44
2
2
1
n(n − 1)
n(n − 1)(n − 2)
= n + 44
+ 64
2!
3!
n
44 n
64 n
=
+
+
1
1
2
1
3
n≥1
n≥2
n≥3
−−−−−−−
n−1
X
0
[2]
n−1
X
(2φi + 1)(2φi + 2)
φT
i+1 =
2
[n = 1, 2, - - -, ∞]
0
=
n
X
1
n(n − 1)
T1 + 2φ(i − 1) = n + (T1 + 2φ − T1 )
2
n−2
X
(n − i)(n − (i + 1))
2
+(2φ)
i
2
1
Substituting a = 1, b =
(2φ + 1)(2φ + 2)
− 1 , c = (2φ)2 , in (5.2)
2
we have
n(n − 1)
(2φ + 1)(2φ + 2)
−1
=n+
2
2
n−2
X
(n − i)(n − (i + 1))
2
+(2φ)
i
2
1
(2φ + 1)(2φ + 2)
n(n − 1)
=n+
−1
2
2!
Triangular Numbers—Some General Theorems · · ·
+(2φ)2
29
n(n − 1)(n − 2)
3!
(2φ + 1)(2φ + 2) − 1 2
n
(2φ)
n
n
2
=
+
+
1
1
3
2
1
n≥1
n≥2
n≥3
n−1
n−1 [3]
X
X
(3i + 1)(3i + 2)
1T
i + 12 =
2
[n = 1, 2, - - -, ∞]
0
0
=
n
X
1
n(n − 1)
T1 + 3(i − 1) = n + (T4 − T1 )
2
n−2
X
(n − i)(n − (i + 1))
2
+3
i
2
1
Substituting a = 1, b = 9, c = 32 = 9, in (5.2) we have
n−2
X
n(n − 1)
(n − i)(n − (i + 1))
=n+9
+9
i
2
2
1
n(n − 1)
n(n − 1)(n − 2)
=n+9
+9
2!
3!
9 n
n
9 n
=
+
+
1
3
1
1
2
n≥3
n≥1
n≥2
n−1 [3]
n−1
X
X
(6i + 1)(6i + 2)
2T
=
i+1
2
0
0
[n = 1, 2, - - -, ∞]
30
=
n
X
1
n(n − 1)
T1 + 6(i − 1) = n + (T7 − T1 )
2
n−2
X
(n − i)(n − (i + 1))
+62
i
2
1
Substituting a = 1, b = 27, c = 62 = 36, in (5.2) we have
n−2
X
(n − i)(n − (i + 1))
n(n − 1)
i
= n + 27
+ 36
2
2
1
n(n − 1)
n(n − 1)(n − 2)
= n + 27
+ 36
2!
3!
27 n
36 n
n
+
+
=
1
2
1
3
1
n≥1
n≥2
n≥3
n−1 [3]
n−1
X
X
(9i + 1)(9i + 2)
3T
=
i+1
2
0
[n = 1, 2, - - -, ∞]
0
=
n
X
1
n(n − 1)
T1 + 9(i − 1) = n + (T10 − T1 )
2
n−2
X
(n − i)(n − (i + 1))
+92
i
2
1
Substituting a = 1, b = 54, c = 92 = 81, in (5.2) we have
n−2
X
n(n − 1)
(n − i)(n − (i + 1))
= n + 54
+ 81
i
2
2
1
n(n − 1)
n(n − 1)(n − 2)
= n + 54
+ 81
2!
3!
Triangular Numbers—Some General Theorems · · ·
31
n
54 n
81 n
=
+
+
1
1
2
1
3
n≥1
n≥2
n≥3
n−1 [3]
n−1
X
X
(12i + 1)(12i + 2)
4T
=
i+1
2
[n = 1, 2, - - -, ∞]
0
0
=
n
X
1
n(n − 1)
T1 + 12(i − 1) = n + (T13 − T1 )
2
n−2
X
(n − i)(n − (i + 1))
2
i
+(12)
2
1
Substituting a = 1, b = 90, c = (12)2 = 144, in (5.2) we have
n−2
X
(n − i)(n − (i + 1))
n(n − 1)
+ 144
i
= n + 90
2
2
1
n(n − 1)(n − 2)
n(n − 1)
+ 144
= n + 90
2!
3!
n
90 n
144 n
=
+
+
1
1
1
2
3
n≥1
n≥2
n≥3
− − − − − − −−
n−1
X
0
[3]
n−1
X
(3φi + 1)(3φi + 2)
φT
i+1 =
2
[n = 1, 2, - - -, ∞]
0
=
n
X
1
T1+3φ(i−1)
n(n − 1)
= n + (T1+3φ − T1 )
2
32
n−2
X
(n − i)(n − (i + 1))
i
+(3φ)
2
1
2
Substituting a = 1, b =
(3φ + 1)(3φ + 2)
− 1 , c = (3φ)2 , in (5.2)
2
we have
(3φ + 1)(3φ + 2)
n(n − 1)
=n+
−1
2
2
n−2
X
(n − i)(n − (i + 1))
2
i
+(3φ)
2
1
n(n − 1)
(3φ + 1)(3φ + 2)
−1
=n+
2
2!
n(n
−
1)(n
−
2)
+(3φ)2
3!
(3φ
+
1)(3φ
+
2)
2
− 1 n
(3φ)
n
n
2
+
+
=
1
3
2
1
1
n≥1
n≥2
n≥3
n−1 [4]
n−1
X
X
(4i + 1)(4i + 2)
1T
i+1 =
2
0
[n = 1, 2, - - -, ∞]
0
=
n
X
1
T1+4(i−1)
n(n − 1)
= n + (T5 − T1 )
2
n−2
X
(n − i)(n − (i + 1))
2
+4
i
2
1
Substituting a = 1, b = 14, c = 42 = 16, in (5.2) we have
Triangular Numbers—Some General Theorems · · ·
33
n−2
X
(n − i)(n − (i + 1))
n(n − 1)
i
+ 16
= n + 14
2
2
1
n(n − 1)
n(n − 1)(n − 2)
= n + 14
+ 16
2!
3!
n
14 n
16 n
=
+
+
1
1
2
1
3
n≥1
n≥2
n≥3
n−1
n−1 [4]
X
X
(8i + 1)(8i + 2)
2T
i+1 =
2
[n = 1, 2, - - -, ∞]
0
0
=
n
X
1
n(n − 1)
T1 + 8(i − 1) = n + (T9 − T1 )
2
n−2
X
(n − i)(n − (i + 1))
2
+8
i
2
1
Substituting a = 1, b = 44, c = (8)2 = 64, in (5.2) we have
n−2
X
(n − i)(n − (i + 1))
n(n − 1)
= n + 44
+ 64
i
2
2
1
n(n − 1)(n − 2)
n(n − 1)
= n + 44
+ 64
2!
3!
44 n
64 n
n
+
+
=
1
1
2
3
1
n≥1
n≥2
n≥3
n−1 [4]
n−1
X
X
(12i + 1)(12i + 2)
3T
=
i+1
2
0
0
[n = 1, 2, - - -, ∞]
34
=
n
X
T1+12(i−1)
1
n(n − 1)
= n + (T13 − T1 )
2
n−2
X
(n − i)(n − (i + 1))
i
+(12)2
2
1
Substituting a = 1, b = 90, c = (12)2 = 144, in (5.2) we have
n−2
X
(n − i)(n − (i + 1))
n(n − 1)
i
+ 144
= n + 90
2
2
1
n(n − 1)
n(n − 1)(n − 2)
= n + 90
+ 144
2!
3!
144 n
n
90 n
=
+
+
1
3
1
1
2
n≥1
n≥2
n≥3
n−1 [4]
n−1
X
X
(16i + 1)(16i + 2)
4T
=
i+1
2
0
[n = 1, 2, - - -, ∞]
0
=
n
X
1
n(n − 1)
T1 + 16(i − 1) = n + (T17 − T1 )
2
n−2
X
(n − i)(n − (i + 1))
+(16)2
i
2
1
Substituting a = 1, b = 152, c = (16)2 = 256, in (5.2) we have
n−2
X
(n − i)(n − (i + 1))
n(n − 1)
+ 256
i
= n + 152
2
2
1
Triangular Numbers—Some General Theorems · · ·
35
n(n − 1)(n − 2)
n(n − 1)
+ 256
= n + 152
2!
3!
n
256 n
152 n
=
+
+
1
1
3
1
2
n≥3
n≥2
n≥1
− − − − − − −−
n−1
X
0
[4]
n−1
X
(φ4i + 1)(φ4i + 2)
φT
i+1 =
2
0
=
n
X
1
[n = 1, 2, - - -, ∞]
n(n − 1)
T1 + 4φ(i − 1) = n + (T1 + 4φ − T1 )
2
n−2
X
(n − i)(n − (i + 1))
2
+(4φ)
i
2
1
Substituting a = 1, b =
(4φ + 1)(4φ + 2)
− 1 , c = (4φ)2 , in (5.2)
2
we have
(4φ + 1)(4φ + 2)
n(n − 1)
=n+
−1
2
2
n−2
X
(n − i)(n − (i + 1))
2
+(4φ)
i
2
1
(4φ + 1)(4φ + 2)
n(n − 1)
=n+
−1
2
2!
2 n(n − 1)(n − 2)
+(4φ)
3!
36
(4φ + 1)(4φ + 2) − 1 2
n
(4φ)
n
n
2
=
+
+
1
1
3
2
1
n≥1
n≥2
n≥3
−−−−−−−
n−1 [R]
n−1
X
X
(Ri + 1)(Ri + 2)
1T
=
i+1
2
[n = 1, 2, - - -, ∞]
0
0
=
n
X
T1+R(i−1)
1
n−2
X
n(n − 1)
(n − i)(n − (i + 1))
2
= n + (T1+R − T1 )
i
+R
2
2
1
Substituting a = 1, b =
(R + 1)(R + 2)
− 1 , c = R2 , in (5.2) we
2
have
(R + 1)(R + 2)
n(n − 1)
=n+
−1
2
2
n−2
X
(n − i)(n − (i + 1))
2
+R
i
2
1
(R + 1)(R + 2)
n(n − 1)
=n+
−1
2
2!
+R
2
n(n − 1)(n − 2)
3!
2 (R + 1)(R + 2) − 1 R
n
n
n
2
+
+
=
1
3
1
2
1
n≥1
n≥2
n≥3
Triangular Numbers—Some General Theorems · · ·
n−1 [R]
n−1
X
X
(2Ri + 1)(2Ri + 2)
2T
=
i+1
2
37
[n = 1, 2, - - -, ∞]
0
0
=
n
X
T1+2R(i−1)
1
n(n − 1)
= n + (T1+2R − T1 )
2
n−2
X
(n − i)(n − (i + 1))
2
i
+(2R)
2
1
(2R + 1)(2R + 2)
− 1 , c = (2R)2 , in (5.2)
Substituting a = 1, b =
2
we have
(2R + 1)(2R + 2)
n(n − 1)
=n+
−1
2
2
n−2
X
(n − i)(n − (i + 1))
2
+(2R)
i
2
1
(2R + 1)(2R + 2)
n(n − 1)
=n+
−1
2
2!
2
+(2R)
n(n − 1)(n − 2)
3!
(2R + 1)(2R + 2) − 1 2
(2R)
n
n
n
2
=
+
+
1
3
1
2
1
n≥3
n≥2
n≥1
n−1 [R]
n−1
X
X
(3Ri + 1)(3Ri + 2)
3T
=
i+1
2
0
[n = 1, 2, - - -, ∞]
0
=
n
X
1
T1+3R(i−1)
n(n − 1)
= n + (T1+3R − T1 )
2
38
n−2
X
(n − i)(n − (i + 1))
i
+(3R)
2
1
2
(3R + 1)(3R + 2)
Substituting a = 1, b =
− 1 , c = (3R)2 , in (5.2)
2
we have
(3R + 1)(3R + 2)
n(n − 1)
=n+
−1
2
2
n−2
X
(n − i)(n − (i + 1))
2
i
+(3R)
2
1
(3R + 1)(3R + 2)
n(n − 1)
=n+
−1
2
2!
+(3R)2
n(n − 1)(n − 2)
3!
(3R
+
1)(3R
+
2)
− 1 n (3R)2 n
n
2
+
+
=
1
3
1
2
1
n≥1
n≥2
n≥3
n−1 [R]
n−1
X
X
(4Ri + 1)(4Ri + 2)
4T
i+1 =
2
0
[n = 1, 2, - - -, ∞]
0
=
n
X
1
T1+4R(i−1)
n(n − 1)
= n + (T1+4R − T1 )
2
n−2
X
(n − i)(n − (i + 1))
2
+(4R)
i
2
1
(4R + 1)(4R + 2)
Substituting a = 1, b =
− 1 , c = (4R)2 , in (5.2)
2
we have
Triangular Numbers—Some General Theorems · · ·
39
(4R + 1)(4R + 2)
n(n − 1)
=n+
−1
2
2
n−2
X
(n − i)(n − (i + 1))
+(4R)2
i
2
1
(4R + 1)(4R + 2)
n(n − 1)
=n+
−1
2
2!
+(4R)2
n(n − 1)(n − 2)
3!
(4R
+
1)(4R
+
2)
2
− 1 n
n
(4R)
n
2
+
+
=
1
1
3
2
1
n≥1
n≥2
n≥3
−−−−−−−
n−1
X
0
[R]
n−1
X
(φRi + 1)(φRi + 2)
φT
=
i+1
2
0
=
n
X
1
[n = 1, 2, - - -, ∞]
n(n − 1)
= n + (T1+φR − T1 )
2
n−2
X
(n − i)(n − (i + 1))
+(φR)2
i
2
1
T1+φR(i−1)
(φR + 1)(φR + 2)
Substituting a = 1, b =
− 1 , c = (φR)2 , in (5.2)
2
we have
(φR + 1)(φR + 2)
n(n − 1)
=n+
−1
2
2
n−2
X (n − i)(n − (i + 1))
+(φR)2
i
2
1
40
(φR + 1)(φR + 2)
n(n − 1)
=n+
−1
2
2!
2
+(φR)
n(n − 1)(n − 2)
3!
(φR + 1)(φR + 2) − 1 2
n
(φR)
n
n
2
+
+
=
1
1
3
2
1
n≥1
n≥2
n≥3
6
Some Interesting Properties of Triangular Numbers
We have proved Theorem 1.1 that
If T = {T0 , T1 , T2 .T3 , - - -, Tn , - - -, ∞}[T0 = 0] are Triangular Numbers
then (2ξ + 1)2 Tn + Tξ [ξ ≥ 0, n ≥ 0] are also Triangular Numbers.
We can utilize this theorem to generate Infinite Inductive Theorems as
follows. The self-evident (though a bit cumbersome) inductive steps of
the Proofs are omitted.
Theorem 6.1. There Exist Infinite Triangular Numbers That Are the sum
of three Triangular Numbers.
Proof. Substituting n = 1 in “(T2ξ + T2ξ+1 )Tn + Tξ = T(3ξn+1) ”[1.1]
we have T2ξ + T2ξ+1 + Tξ = T(3ξ+1) · [ξ ≥ 1] proving the theorem.
Theorem 6.2. There Exist Infinite Triangular Numbers That Are the sum
of Five Triangular Numbers.
Proof. ((2ξ + 1)2 )2 + T(3ξ+1) = (T2ξ + T2ξ+1 )2 + T(3ξ+1)
Triangular Numbers—Some General Theorems · · ·
41
= T(2ξ+1)2 + T[(2ξ+1)2 +1] + T2ξ + T2ξ+1 + Tξ
= T[(2ξ+1)2 +(3ξ+1)] · [ξ ≥ 1] proving the theorem.
Theorem 6.3. There Exist Infinite Triangular Numbers That Are the sum
of Seven Triangular Numbers.
Proof. (((2ξ + 1)2 )2 )2 + T[(2ξ+1)2 +(3ξ+1)] =
T[((2ξ+1)2 )2 ] + T[((2ξ+1)2 )2 +1] +
T(2ξ+1)2 + T[(2ξ+1)2 +1] + T2ξ + T2ξ+1 + Tξ
= T[((2ξ+1)2 )2 +(2ξ+1)2 +(3ξ+1)] .
[ξ ≥ 1]
proving the theorem.
Theorem 6.4. There Exist Infinite Triangular Numbers That Are the sum
of Nine Triangular Numbers.
Proof. ((((2ξ + 1)2 )2 )2 )2 + T[((2ξ+1)2 )2 +(2ξ+1)2 +(3ξ+1)] =
T[(((2ξ+1)2 )2 )2 ] + T[(((2ξ+1)2 )2 )2 +1] + T[((2ξ+1)2 )2 ] +
T[((2ξ+1)2 )2 ]+1] + T(2ξ+1)2 + T[(2ξ+1)2 +1] + T2ξ + T2ξ+1 + Tξ
= T[(((2ξ+1)2 )2 )2 +((2ξ+1)2 )2 +(2ξ+1)2 +(3ξ+1)] ·
[ξ ≥ 1]
proving the theorem.
42
Theorem 6.τ [τ ≥ 1]. There Exist Infinite Triangular Numbers That
Are the sum of (2τ + 1), [τ ≥ 1] Triangular Numbers.
We continue the Inductive Proof for [τ ≥ 3]
((((2ξ + 1)2 )2 )2 )τ times +
T[(((2ξ+1)2 )2 ))))(τ −2)times - - -+((2ξ+1)2 )2 +(2ξ+1)2 +(3ξ+1)] =
T[((((2ξ+1)2 )2 )2 )(τ −1)times ] + T[((((2ξ+1)2 )2 )2 )(τ −1)times +1] +
T[((((2ξ+1)2 )2 )2 )(τ −2)times ] + T[((((2ξ+1)2 )2 )2 )(τ −2)times +1] +
T[((((2ξ+1)2 )2 )2 )(τ −3)times ] + T[((((2ξ+1)2 )2 )2 )(τ −3)times +1] +
−−−−−−−
T[(((2ξ+1)2 )2 )] + T[((2ξ+1)2 )2 )+1] +
T(2ξ+1)2 + T[(2ξ+1)2 +1] + T2ξ + T2ξ+1 + Tξ
= T[((((2ξ+1)2 )2 )2 )))(τ −1)times + - - -+((2ξ+1)2 )2 + (2ξ + 1)2 + (3ξ + 1)]·
[ξ ≥ 1][τ ≥ 3]
Simplifying
τ
(2ξ + 1)2 + T[(2ξ+1)2(τ −2) + - - -+(2ξ+1)4 +(2ξ+1)2 +(3ξ+1)] =
T[(2ξ+1)2(τ −1) ] + T[(2ξ+1)2(τ −1) +1] +
T[(2ξ+1)2(τ −2) ] + T[(2ξ+1)2(τ −2) +1] +
Triangular Numbers—Some General Theorems · · ·
43
T[(2ξ+1)2(τ −3) ] + T[(2ξ+1)2(τ −3) +1] +
−−−−−−−
T[(2ξ+1)4 ] + T[(2ξ+1)4 +1] + T(2ξ+1)2 + T[(2ξ+1)2 +1]
+T2ξ + T2ξ+1 + Tξ
= T[(2ξ+1)2(τ −1) + - - -+(2ξ+1)4 +(2ξ+1)2 +(3ξ+1)] · · ·
[ξ ≥ 1][τ ≥ 3]
proving the theorems.
We have Proved Theorem 1.2 that
If T = {T0 , T1 , T2 , T3 , - - -Tn , - - -, - - -∞}[T0 = 0] are Triangular
Numbers then (2λ)2 Tn + Tλ + λn[λ ≥ 1, n ≥ 0] are also Triangular
Numbers.
We can utilise this theorem to generate Infinite Inductive Theorems as
follows. The self-evident (though a bit cumbersome) inductive steps of
the Proofs are omitted.
Theorem 6.5. There Exist Infinite Triangular Numbers That Are the sum
of three Triangular Numbers and a specified constant λ = 1, 2, 3, - - -∞
or Equivalently.
For each λ = 1, 2, 3, - - -, ∞ There Exist Three Triangular Numbers,
the Sum of which and λ is yet another Triangular Number.
Proof. Substituting n = 1 in “(T2λ−1 + T2λ )Tn + Tλ + λn = T3λn ”
[1.2] we have T2λ−1 +T2λ +Tλ +λ = T3λ
[λ ≥ 1] proving the theorem.
44
When λ is a Triangular Number we have the Infinite cases when we
have Triangular Numbers that are the Sum of Four Triangular Numbers
[n ≥ 1]
T[n(n+1)−1] + Tn(n+1) + T [n(n+1)] + Tn = T 3n(n+1)
2
2
proving the theorem.
Theorem 6.6. There Exist Infinite Triangular Numbers That Are the sum
of Five Triangular Numbers and a specified constant λ = 1, 2, 3, - - -, ∞
or Equivalently.
For each λ = 1, 2, 3, - - -, ∞ There Exist Five Triangular Numbers,
the Sum of which and λ is yet another Triangular Number.
Proof. ((2λ)2 )2 + T3λ = (T2λ−1 + T2λ )2 + T3λ
= T(2λ)2 + T[(2λ)2 +1] + T2λ−1 + T2λ + Tλ + λ
= T[(2λ)2 +(3λ)] ·
[λ ≥ 1] proving the theorem.
When λ is a Triangular Number we have the Infinite cases when we have
Triangular Numbers that are the Sum of Six Triangular Numbers
T[(n(n+1))2 ] + T[(n(n+1))2 +1] + T[n(n+1)−1] + Tn(n+1) + T [n(n+1)] + Tn
2
= T[(n(n+1))2 + 3n(n+1) ] ·
[n ≥ 1]
2
proving the theorem.
Theorem 6.7. There Exist Infinite Triangular Numbers That Are the sum
of Seven Triangular Numbers and a specified constant λ = 1, 2, 3, - - -, ∞
or Equivalently.
Triangular Numbers—Some General Theorems · · ·
45
For each λ = 1, 2, 3, - - -, ∞ There Exist Seven Triangular Numbers,
the Sum of which and λ is yet another Triangular Number.
Proof.
(((2λ)2 )2 )2 + T[(2λ)2 +(3λ)] = T[((2λ)2 )2 ] + T[((2λ)2 )2 +1]
+ T(2λ)2 + T[(2λ)2 +1] + T2λ−1 + T2λ + Tλ + λ
= T[((2λ)2 )2 +(2λ)2 +(3λ)] ·
[λ ≥ 1]
proving the theorem.
When λ is a Triangular Number we have the Infinite cases when we have
Triangular Numbers that are the Sum of Eight Triangular Numbers
T[((n(n+1))2 )2 ] + T[((n(n+1))2 )2 +1] + T[(n(n+1))2 ] +
T[(n(n+1))2 +1] + T[(n(n+1))−1] + T(n(n+1)) + T [n(n+1)] + Tn
2
= T[((n(n+1))2 )2 +(n(n+1))2 + 3n(n+1) ] ·
[n ≥ 1]
2
proving the theorem.
Theorem 6.8. There Exist Infinite Triangular Numbers That Are the
sum of Nine Triangular Numbers and a specified constant λ = 1, 2, 3,
- - -, ∞ or Equivalently.
For each λ = 1, 2, 3, - - -, ∞ There Exist Nine Triangular Numbers, the
Sum of which and λ is yet another Triangular Number.
Proof. ((((2λ)2 )2 )2 )2 + T[((2λ)2 )2 +(2λ)2 +(3λ)] =
46
T[(((2λ)2 )2 )2 ] + T[(((2λ)2 )2 )2 +1] + T[((2λ)2 )2 ] + T[((2λ)2 )2 +1] +
T(2λ)2 + T[(2λ)2 +1] + T2λ−1 + T2λ + Tλ + λ
= T[(((2λ)2 )2 )2 +((2λ)2 )2 +(2λ)2 +(3λ)] ·
[λ ≥ 1]
proving the theorem.
When λ is a Triangular Number we have the Infinite cases when we have
Triangular Numbers that are the Sum of Ten Triangular Numbers
T[(((n(n+1))2 )2 )2 ] + T[(((n(n+1))2 )2 )2 +1]
T[((n(n+1))2 )2 ] + T[((n(n+1))2 )2 +1]
T[(n(n+1))2 ] + T[(n(n+1))2 +1] + T[n(n+1)−1] + Tn(n+1) + T [n(n+1)] + Tn
2
= T[(((n(n+1))2 )2 )2 +((n(n+1))2 )2 +(n(n+1))2 + 3n(n+1) ] · [n ≥ 1]
2
proving the theorem.
− − − − −−
Theorem 6. ψ[ψ ≥ 1]. There Exist Infinite Triangular Numbers That
Are the sum of (2ψ + 1) [ψ ≥ 1] Triangular Numbers and a specified
constant λ = 1, 2, 3, - - -, ∞ Or Equivalently.
For each λ = 1, 2, 3, - - -, ∞ There Exist (2ψ + 1)[ψ ≥ 1] Triangular
Numbers, the Sum of which and λ is yet another Triangular Number.
We continue the Inductive Proof for [ψ ≥ 3]
Triangular Numbers—Some General Theorems · · ·
47
((((2λ)2 )2 )2 ))))ψ times +
T[(((2λ)2 )2 ))))(ψ−2) times - - -+((2λ)2 )2 (2λ)2 +(3λ)] =
T[((((2λ)2 )2 )2 ))(ψ−1) times ] + T[((((2λ)2 )2 )2 ))(ψ−1) times +1] +
T[((((2λ)2 )2 )2 ))(ψ−2) times ] + T[((((2λ)2 )2 )2 ))(ψ−2) times +1] +
T[((((2λ)2 )2 )2 ))(ψ−3) times ] + T[((((2λ)2 )2 )2 ))(ψ−3) times +1] +
− − − − − − −−
T[((2λ)2 )2 ] + T[((2λ)2 )2 +1] + T(2λ)2 + T[(2λ)2 +1] + T2λ−1 + T2λ + Tλ + λ
= T[((((2λ)2 )2 )2 )(ψ−1) times + - - -+((2λ)2 )2 +(2λ)2 +(3λ)] ·
[λ ≥ 1]
[ψ ≥ 3]
Simplifying
ψ
(2λ)2 + T[(2λ)2ψ−2 + - - -+(2λ)4 +(2λ)2 +(3λ)] =
T[(2λ)2(ψ−1) ] + T[(2λ)2(ψ−1) +1] +
T[(2λ)2(ψ−2) ] + T[(2λ)2(ψ−2) +1] +
T[(2λ)2(ψ−3) ] + T[(2λ)2( ψ−3) +1] +
−−−−−−−−−
T[(2λ)4 ] + T[(2λ)4 +1] + T(2λ)2 + T[(2λ)2 +1] + T2λ−1 + T2λ + Tλ + λ
= T[(2λ)2(ψ−1) + - - -+(2λ)4 +(2λ)2 +(3λ)] ·
proving the theorems.
[λ ≥ 1][ψ ≥ 3]
48
When λ is a Triangular Number we have the Infinite cases when we have
Triangular Numbers that are the Sum of (2ψ + 2) = 2(ψ + 1)[ψ ≥ 1]
Triangular Numbers.
We continue the Inductive Proof for [ψ ≥ 3]
T[(n(n+1))2(ψ−1) ] + T[(n(n+1))2(ψ−1) +1] +
T[(n(n+1))2(ψ−2) ] + T[(n(n+1))2(ψ−2) +1] +
T[(n(n+1))2(ψ−3) ] + T[(n(n+1))2(ψ−3) +1] +
− − − − − − −−
T[(n(n+1))4 ] + T[(n(n+1))4 +1] +
T[(n(n+1))2 ] + T[(n(n+1))2 +1] + T[n(n+1)2 −1] + Tn(n+1) + T [n(n+1)] + Tn
2
= T[(n(n+1))2(ψ−1) + - - -+(n(n+1))4 +(n(n+1))2 + 3n(n+1) ] · [n ≥ 1]
2
proving the theorems.
This Proves the General theorem
Theorem 6.9. For any Natural Number N ≥ 2, there Exist Infinite
Triangular Numbers that are the Sum of N Triangular Numbers each.
Proof. For N = 2 we have the standard result. (Ref. 6)
For N > 2, we have proved the result.
For N = 3, 5, 7, - - - and N = 4, 6, 8, - - - in Theorems 6) and Theorems
6a) respectively. This Completes the Proof.
Triangular Numbers—Some General Theorems · · ·
7
49
Theorems on two special classes of Triangular Numbers
Theorem 7.1. If x =
n(n + 2α + 1)
n(n + 2α + 1)
+1, y = n+1, z =
2
2
Tx = Ty + Tz + (α − 1)n = Ty + Tz + (α − 1)(y − 1)
[α = 0, 1, 2, - - -, α]
Proof. If x =
7–1–1)
[n = 0, 1, 2, - - -, α]
n(n + 2α + 1)
n(n + 2α + 1)
+ 1, y = n + 1, z =
2
2
For, α = 0 we can prove by inducting on “n”
n(n + 1)
n(n + 1)
+ 1, y = n + 1, z =
2
2
Tx = Ty + Tz − n = Ty + Tz − (y − 1) [n = 0, 1, 2, - - -, ∞]
x=
7–1–2)
For, α = 1 we can prove by inducting on “n”
n(n + 3)
n(n + 3)
+ 1, y = n + 1, z =
2
2
Tx = Ty + Tz [n = 0, 1, 2, - - -, ∞]
x=
which is the standard result that states that there exist infinite Triangular
Numbers that are the sum of two other Triangular Numbers. [Ref. 6]
7–1–3)
For, α = 2 we can prove by inducting on “n”
n(n + 5)
n(n + 5)
+ 1, y = n + 1, z =
2
2
Tx = Ty + Tz + n = Ty + Tz + (y − 1) [n = 0, 1, 2, - - -, ∞]
x=
50
7–1–4)
For, α = 3 we can prove by inducting on “n”
n(n + 7)
n(n + 7)
+ 1, y = n + 1, z =
2
2
Tx = Ty + Tz + n = Ty + Tz + (y − 1) [n = 0, 1, 2, - - -, ∞]
x=
7–1–5)
For, α = 4 we can prove by inducting on “n”
n(n + 9)
n(n + 9)
+ 1, y = n + 1, z =
2
2
Tx = Ty + Tz + n = Ty + Tz + (y − 1) [n = 0, 1, 2, - - -, ∞]
x=
7–1–6) For, α = 5 we can prove by inducting on “n”
n(n + 11)
n(n + 11)
+ 1, y = n + 1, z =
2
2
Tx = Ty + Tz + n = Ty + Tz + (y − 1) [n = 0, 1, 2, - - -, ∞]
x=
7–1–k+1) For, α = k we can prove by inducting on “n”
n(n + 2k + 1)
n(n + 2k + 1)
+ 1, y = n + 1, z =
2
2
Tx = Ty + Tz + (k − 1)n = Ty + Tz + (k − 1)(y − 1)
x=
[k = 0, 1, 2, · · · , ∞] [n = 0, 1, 2, · · · , ∞]
7–1–k+2) For, α = k + 1 we can prove by inducting on “n”
x=
n(n + 2(k + 1) + 1)
+ 1,
2
y = n + 1, z =
n(n + 2(k + 1) + 1)
2
Triangular Numbers—Some General Theorems · · ·
Tx = Ty + Tz + (k)n = Ty + Tz + (k)(y − 1)
[k = 0, 1, 2, · · · , ∞] [n = 0, 1, 2, · · · , ∞]
thus Completing the Induction on “α” proving the theorem that, if
n(n + 2α + 1)
n(n + 2α + 1)
+ 1, y = n + 1, z =
2
2
Tx = Ty + Tz + (α − 1)n = Ty + Tz + (α − 1)(y − 1)
x=
[α = 0, 1, 2, · · · , ∞]
[n = 0, 1, 2, · · · , ∞]
Theorem 7.2. If
(n + 1)(n + 2β)
(n + 1)(n + 2β)
+ 1, y = n + 1, z =
2
2
Tx = Ty + Tz + β + (β − 1)n = Ty + Tz + β + (β − 1)(y − 1)
x=
[β = 1, 2, 3, · · · , ∞] [n = 0, 1, 2, · · · , ∞]
Proof. If
x=
(n + 1)(n + 2β)
(n + 1)(n + 2β)
+ 1, y = n + 1, z =
.
2
2
7–2–1) For, β = 1 we can prove by inducting on “n”
(n + 1)(n + 2)
(n + 1)(n + 2)
+ 1, y = n + 1, z =
2
2
Tx = Ty + Tz + 1 [n = 0, 1, 2, - - -, ∞]
x=
51
52
7–2–2) For, β = 2 we can prove by inducting on “n”
(n + 1)(n + 4)
(n + 1)(n + 4)
+ 1, y = n + 1, z =
2
2
Tx = Ty + Tz + 2 + n = Ty + Tz + 2 + (y − 1) [n = 0, 1, 2, - - -, ∞]
x=
7–2–3) For, β = 3 we can prove by inducting on “n”
(n + 1)(n + 6)
(n + 1)(n + 6)
+ 1, y = n + 1, z =
2
2
Tx = Ty + Tz + 3 + 2n = Ty + Tz + 3 + 2(y − 1) [n = 0, 1, 2, - - -, ∞]
x=
7–2–4) For, β = 4 we can prove by inducting on “n”
(n + 1)(n + 8)
(n + 1)(n + 8)
+ 1, y = n + 1, z =
2
2
Tx = Ty + Tz + 4 + 3n = Ty + Tz + 4 + 3(y − 1) [n = 0, 1, 2, · · · , ∞]
x=
7–2–5) For, β = 5 we can prove by inducting on “n”
(n + 1)(n + 10)
(n + 1)(n + 10)
+ 1, y = n + 1, z =
2
2
Tx = Ty + Tz + 5 + 4n = Ty + Tz + 5 + 4(y − 1) [n = 0, 1, 2, · · · , ∞]
x=
−−−−−−−
Triangular Numbers—Some General Theorems · · ·
53
7–2–k) For, β = k we can prove by inducting on “n”
(n + 1)(n + 2k)
(n + 1)(n + 2k)
+ 1, y = n + 1, z =
2
2
Tx = Ty + Tz + k + (k − 1)n = Ty + Tz + k + (k − 1)(y − 1)
x=
[k = 1, 2, 3, · · · , ∞] [n = 0, 1, 2, · · · , ∞]
7–2–k) For, β = k + 1 we can prove by inducting on “n”
(n + 1)(n + 2(k + 1))
(n + 1)(n + 2(k + 1))
+ 1, y = n + 1, z =
2
2
Tx = Ty + Tz + (k + 1) + (k)n = Ty + Tz + (k + 1) + (k)(y − 1)
x=
[k = 1, 2, 3, · · · , ∞] [n = 0, 1, 2, · · · , ∞]
thus completing the induction on “β ” proving the theorem that
If
(n + 1)(n + 2β)
(n + 1)(n + 2β)
+ 1, y = n + 1, z =
2
2
Tx = Ty + Tz + β + (β − 1)n = Ty + Tz + β + (β − 1)(y − 1)
x=
[β = 1, 2, 3 - - -, ∞] [n = 0, 1, 2, - - -, ∞]
We come to the end of this paper. The Super-Sums of each Sub-Class
of Triangular Numbers upto any Level and other interesting things are
possible. [Ref. 7,8,9.]
54
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(1707 –1783)
[]
[] = from 6)
6) David M. Burton:
ELEMENTARY NUMBER THEORY.
UNIVERSAL BOOK STALL. NEW DELHI. Second Edition. Reprint
1998. (p. 18–19)
7) Narayanan Raghunathan
Generalised arithmetic progressions – a survey [ Submitted ]
8) Narayanan Raghunathan
Generalised geometric progressions – a survey [ Submitted ]
9) Narayanan Raghunathan.
THE UNIVERSES OF TRIANGULAR NUMBER RHYTHMS. [Getting Ready]