Editing Triangular number (section)

==Relations to other figurate numbers==
Triangular numbers have a wide variety of relations to other figurate numbers.

Most simply, the sum of two consecutive triangular numbers is a square number, since:<ref>{{cite journal
| last1 = Beldon
| first1 = Tom
| last2 = Gardiner
| first2 = Tony
| title = Triangular Numbers and Perfect Squares
| journal = The Mathematical Gazette
| volume = 86
| issue = 507
| year = 2002
| pages = 423–431
| doi = 10.2307/3621134
| url = https://doi.org/10.2307/3621134
| access-date = 25 April 2024
| jstor = 3621134
}}</ref><ref>{{cite web |title=Triangular Number |url=https://mathworld.wolfram.com/TriangularNumber.html |author=Eric W. Weisstein |publisher=Wolfram MathWorld |accessdate=2024-04-14}} See equations 18 - 20.</ref>

:<math>T_{n - 1} + T_{n}</math>
:<math>= \frac{1}{2} \, n(n-1) + \frac{1}{2} \, n (n + 1)</math>
:<math>= \frac{1}{2} \, n\Bigl((n - 1) + (n + 1)\Bigr)</math>
:<math>= n^2</math> 

with the sum being the square of the difference between the two (and thus the difference of the two being the square root of the sum):
<math display="block">T_n + T_{n-1} = \left (\frac{n^2}{2} + \frac{n}{2}\right) + \left(\frac{\left(n-1\right)^2}{2} + \frac{n-1 \vphantom{\left(n-1\right)^2}}{2} \right ) = \left (\frac{n^2}{2} + \frac{n}{2}\right) + \left(\frac{n^2}{2} - \frac{n}{2} \right ) = n^2 = (T_n - T_{n-1})^2.</math>

This property, colloquially known as the theorem of [[Theon of Smyrna]],<ref>{{cite book
    | last1 = Shell-Gellasch
    | first1 = Amy
    | last2 = Thoo
    | first2 = John
    | author-link = 
    | date = October 15, 2015
    | title = Algebra in Context: Introductory Algebra from Origins to Applications
    | url = https://doi.org/10.1353/book.49475
    | location = 
    | publisher = Johns Hopkins University Press
    | page = 210
    | doi = 10.1353/book.49475
 | isbn = 9781421417288
}}</ref> is visually demonstrated in the following sum, which represents <math>T_{4} + T_{5} = 5^2</math> as [[Digit sum|digit sums]]:

<math>\begin{array}{ccccccc}
    & 4 & 3 & 2 & 1 & \\
+   & 1 & 2 & 3 & 4 & 5 \\
\hline
    & 5 & 5 & 5 & 5 & 5
\end{array}</math>

This fact can also be demonstrated graphically by positioning the triangles in opposite directions to create a square:
{{bi|left=1.6
|6 + 10 {{=}} 16 &nbsp;&nbsp;&nbsp; [[File:Square number 16 as sum of two triangular numbers.svg]] &nbsp;&nbsp;&nbsp; 10 + 15 {{=}} 25 &nbsp;&nbsp;&nbsp; [[File:Square number 25 as sum of two triangular numbers.svg]]
}}

The double of a triangular number, as in the visual proof from the above section {{section link||Formula}}, is called a [[pronic number]].

There are infinitely many triangular numbers that are also square numbers; e.g., 1, 36, 1225. Some of them can be generated by a simple recursive formula:
<math display=block>S_{n+1} = 4S_n \left( 8S_n + 1\right)</math> with <math>S_1 = 1.</math>

''All'' [[square triangular number]]s are found from the recursion
<math display=block>S_n = 34S_{n-1} - S_{n-2} + 2</math> with <math>S_0 = 0</math> and <math>S_1 = 1.</math>

[[File:Nicomachus_theorem_3D.svg|thumb|A square whose side length is a triangular number can be partitioned into squares and half-squares whose areas add to cubes. This shows that the square of the {{mvar|n}}th triangular number is equal to the sum of the first {{mvar|n}} cube numbers.]]

Also, the [[squared triangular number|square of the {{mvar|n}}th triangular number]] is the same as the sum of the cubes of the integers 1 to {{mvar|n}}. This can also be expressed as
<math display="block"> \sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k \right)^2.</math>

The sum of the first {{mvar|n}} triangular numbers is the {{mvar|n}}th [[tetrahedral number]]:
<math display=block> \sum_{k=1}^n T_k = \sum_{k=1}^n \frac{k(k+1)}{2} = \frac {n(n+1)(n+2)} {6}.</math>

More generally, the difference between the {{mvar|n}}th [[polygonal number|{{mvar|m}}-gonal number]] and the {{mvar|n}}th {{math|(''m'' + 1)}}-gonal number is the {{math|(''n'' − 1)}}th triangular number. For example, the sixth [[heptagonal number]] (81) minus the sixth [[hexagonal number]] (66) equals the fifth triangular number, 15. Every other triangular number is a hexagonal number. Knowing the triangular numbers, one can reckon any [[centered polygonal number]]; the {{mvar|n}}th centered {{mvar|k}}-gonal number is obtained by the formula
<math display=block>Ck_n = kT_{n-1}+1</math>

where {{mvar|T}} is a triangular number.

The positive difference of two triangular numbers is a [[trapezoidal number]].

The pattern found for triangular numbers <math> \sum_{n_1=1}^{n_2}n_1=\binom{n_2+1}{2}</math> and for tetrahedral numbers <math> \sum_{n_2=1}^{n_3}\sum_{n_1=1}^{n_2} n_1=\binom{n_3+2}{3},</math> which uses [[binomial coefficient]]s, can be generalized. This leads to the formula:<ref>{{Cite journal| last=Baumann|first=Michael Heinrich |date=2018-12-12|title=Die {{mvar|k}}-dimensionale Champagnerpyramide |journal=Mathematische Semesterberichte |language=de|volume=66|pages=89–100 |doi=10.1007/s00591-018-00236-x |s2cid=125426184 |issn=1432-1815 |url=https://epub.uni-bayreuth.de/3850/1/Baumann_Champagnerpyramide.pdf}}</ref>
<math display=block> \sum_{n_{k-1}=1}^{n_k}\sum_{n_{k-2}=1}^{n_{k-1}} \dots \sum_{n_2=1}^{n_3}\sum_{n_1=1}^{n_2}n_1 =\binom{n_k+k-1}{k}</math>
[[File:tetrahedral_triangular_number_10.svg|thumb|The fourth triangular number equals the third tetrahedral number as the ''n''th ''k''-simplex number equals the ''k''th ''n''-simplex number due to the symmetry of [[Pascal's triangle]], and its diagonals being simplex numbers; similarly, the fifth triangular number (15) equals the third [[pentatope number]], and so forth]]