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[[File:TrianguloPascal.jpg|thumb|right|upright=1.25|[[Blaise Pascal|Pascal]]'s version of the triangle]]
The pattern of numbers that forms Pascal's triangle was known well before Pascal's time. The Persian mathematician [[Al-Karaji]] (953–1029) wrote a now-lost book which contained the first description of Pascal's triangle.<ref>{{Cite book|url=https://books.google.com/books?id=kt9DIY1g9HYC&q=al+karaji+pascal%27s+triangle&pg=PA132|title=Encyclopaedia of the History of Science, Technology, and Medicine in Non-Western Cultures|last=Selin|first=Helaine|date=2008-03-12|publisher=Springer Science & Business Media|isbn=9781402045592|language=en|page=132|bibcode=2008ehst.book.....S|quote=Other, lost works of al-Karaji's are known to have dealt with inderterminate algebra, arithmetic, inheritance algebra, and the construction of buildings. Another contained the first known explanation of the arithmetical (Pascal's) triangle; the passage in question survived through al-Sama'wal's Bahir (twelfth century) which heavily drew from the Badi.}}</ref><ref>{{Cite book |last=Rashed |first=R. |url=https://books.google.com/books?id=vSkClSvU_9AC&pg=PA62 |title=The Development of Arabic Mathematics: Between Arithmetic and Algebra |date=1994-06-30 |publisher=Springer Science & Business Media |isbn=978-0-7923-2565-9 |pages=63 |language=en}}</ref><ref>{{Cite book|url=https://books.google.com/books?id=kAjABAAAQBAJ&q=al+karaji+binomial+theorem&pg=PA54|title=From Alexandria, Through Baghdad: Surveys and Studies in the Ancient Greek and Medieval Islamic Mathematical Sciences in Honor of J.L. Berggren|last1=Sidoli|first1=Nathan|last2=Brummelen|first2=Glen Van|date=2013-10-30|publisher=Springer Science & Business Media|isbn=9783642367366|language=en|page=54|quote=However, the use of binomial coefficients by Islamic mathematicians of the eleventh century, in a context which had deep roots in Islamic mathematics, suggests strongly the table was a local discovery - most probably of al-Karaji."}}</ref> In India, the ''[[Chandaḥśāstra]]'' by the Indian
| last = Coolidge | first = J. L. | author-link = Julian Coolidge
| journal = [[The American Mathematical Monthly]]
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Thus the extreme left and right coefficients remain as 1, and for any given <math>0 < k < n + 1 </math>, the coefficient of the <math>x^{k} </math> term in the polynomial <math>(x + 1)^{n + 1} </math> is equal to <math>a_{k-1} + a_{k} </math>, the sum of the <math>x^{k-1} </math> and <math>x^{k} </math> coefficients in the previous power <math>(x + 1)^n </math>. This is indeed the downward-addition rule for constructing Pascal's triangle.
It is not difficult to turn this argument into a [[proof (mathematics)|proof]] (by [[mathematical induction]]) of the binomial theorem.
Since <math>(a + b)^{n} = b^{n}(\tfrac{a}{b} + 1 )^{n} </math>, the coefficients are identical in the expansion of the general case.
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Pascal's triangle has many properties and contains many patterns of numbers.
[[File:Pascal's Triangle animated binary rows.gif|thumb|upright=1|Each frame represents a row in Pascal's triangle. Each column of pixels is a number in binary with the least significant bit at the bottom. Light pixels represent 1 and dark pixels 0.]]
[[File:pascal_triangle_compositions.svg|thumb|upright=1|The numbers of [[composition (combinatorics)|compositions]] of ''n''
=== Rows ===
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}}.</ref> Specifically, define the sequence <math> s_{n}</math> for all <math>n \ge 0</math> as follows: <math>s_{n} = \prod_{k = 0}^{n} {n \choose k} = \prod_{k = 0}^{n} \frac{n!}{k!(n-k)!}</math> {{pb}} Then, the ratio of successive row products is <math display="block">\frac{s_{n+1}}{s_{n}} =
\frac{ \displaystyle (n+1)!^{n+2} \prod_{k = 0}^{n + 1} \frac{1}{k!^2}}{\displaystyle n!^{n+1}\prod_{k=0}^{n}{\frac{1}{k!^2}}} = \frac{(n + 1)^n}{n!}</math> and the ratio of these ratios is <math display="block">\frac{s_{n + 1} \cdot s_{n - 1}}{s_{n}^{2}} =
\left( \frac{n + 1}{n} \right)^n, ~ n\ge 1.</math> The right-hand side of the above equation takes the form of the limit definition of [[e (mathematical constant)|<math>e</math>]] <math display="block">e =\lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^{n}.</math>
* [[pi|<math>\pi</math>]] can be found in Pascal's triangle by use of the [[Nilakantha Somayaji|Nilakantha]] infinite series.<ref>
| last = Foster | first = T.
| doi = 10.5951/mathteacher.108.4.0246
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===Construction of Clifford algebra using simplices===
Labelling the elements of each n-simplex matches the basis elements of [[Clifford algebra]] used as forms in [[
=== Relation to geometry of polytopes ===
{{more
Each row of Pascal's triangle gives the number of elements (such as edges and corners) of each dimension in a corresponding [[simplex]] (such as a triangle or tetrahedron). In particular, for {{math| ''k'' > 0}}, the {{mvar|k}}th entry in the {{mvar|n}}th row is the number of {{math|(''k'' − 1)}}-dimensional elements in a {{math|(''n'' − 1)}}-dimensional simplex. For example, a triangle (the 2-dimensional simplex) one 2-dimensional element (itself), three 1-dimensional elements (lines, or edges), and three 0-dimensional elements ([[Vertex (graph theory)|vertices]], or corners); this corresponds to the third row 1, 3, 3, 1 of Pascal's triangle. This fact can be explained by combining Pascal's rule for generating the triangle with the geometric construction of simplices: each simplex is formed from a simplex of one lower dimension by the addition of a new vertex, outside the space in which the lower-dimensional simplex lies. Then each {{mvar|d}}-dimensional element in the smaller simplex remains a {{mvar|d}}-dimensional element of the higher simplex, and each {{math|(''d'' − 1)}}-dimensional element when joined to the new vertex forms a new {{mvar|d}}-dimensional element of the higher simplex.<ref>{{Cite book |last=Coxeter |first=Harold Scott Macdonald |url=https://books.google.com/books?id=iWvXsVInpgMC |title=Regular Polytopes |date=1973-01-01 |publisher=Courier Corporation |isbn=978-0-486-61480-9 |edition=3rd |pages=118–144 |language=en |chapter=Chapter VII: ordinary polytopes in higher space, 7.2: Pyramids, dipyramids and prisms}}</ref>
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=== To complex numbers ===
When the factorial function is defined as <math>z! = \Gamma(z + 1)</math>, Pascal's triangle can be extended beyond the integers to <math>\Complex</math>, since <math>\Gamma(z + 1)</math> is [[
| display-authors = etal
| last = Hilton | first = P.
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| quote = But these in the alternate areas, which are given, I observed were the same with the figures of which the several ascending powers of the number 11 consist, viz. <math>11^{0}</math>, <math>11^{1}</math>, <math>11^{2}</math>, <math>11^{3}</math>, <math>11^{4}</math>, etc. that is, first 1; the second 1, 1; the third 1, 2, 1; the fourth 1, 3, 3, 1; the fifth 1, 4, 6, 4, 1, and so on
| url = https://www.loc.gov/item/42048007/
}}.</ref> In 1964, Robert L. Morton presented the more generalized argument that each [[#Rows|row]] <math>n</math> can be read as a radix <math>a</math> numeral, where [[E (mathematical constant)|<math>\lim_{n \to \infty} 11^{n}_{a}</math>]] is the hypothetical terminal row, or [[
| last = Morton | first = Robert L.
| issue = 6
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| last = Kallós | first = Gábor
| issue = 1
| journal = Annales Mathématiques Blaise Pascal
| pages = 1–15
| title = A generalization of Pascal's triangle using powers of base numbers
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| doi = 10.5802/ambp.211
| url = https://ambp.centre-mersenne.org/item/10.5802/ambp.211.pdf
}}.</ref> as demonstrated [[#Binomial expansions|above]]. Thus, when the entries of the row are concatenated and read in radix <math>a</math> they form the numerical equivalent of <math>(a + 1)^{n} = 11^{n}_{a}</math>. If <math>c = a + 1</math> for <math>c < 0</math>, then the theorem [[
| display-authors = etal
| last = Hilton | first = P.
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: <math>11^{12}_{12} = 1:10:56:164:353:560:650:560:353:164:56:10:1_{12} = 27433a9699701_{12}</math>
with compound digits (delimited by ":") in radix twelve. The digits from <math>k = n - 1</math> through <math>k = 1</math> are compound because these row entries compute to values greater than or equal to twelve. To [[
| last = Fjelstad | first = P.
| issue = 9
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| year = 1991
| doi-access = free
}}.</ref> the numeral, simply carry the first compound entry's prefix, that is, remove the prefix of the coefficient <math>\textstyle {n \choose n - 1}</math> from its leftmost digit up to, but excluding, its rightmost digit, and use radix-twelve arithmetic to sum the removed prefix with the entry on its immediate left, then repeat this process, proceeding leftward, until the leftmost entry is reached. In this particular example, the normalized string ends with <math>01</math> for all <math>n</math>. The leftmost digit is <math>2</math> for <math>n > 2</math>, which is obtained by carrying the <math>1</math> of <math>10_{n}</math> at entry <math>k = 1</math>. It follows that the length of the normalized value of <math>11^{n}_{n}</math> is [[Bijection|equal]] to the row length, <math>n + 1</math>. The integral part of <math>1.1^{n}_{n}</math> contains exactly one digit because <math>n</math> (the number of places to the left the decimal has moved) is one less than the row length. Below is the normalized value of <math>1.1^{1234}_{1234}</math>. Compound digits remain in the value because they are radix <math>1234</math> [[
: <math>1.1^{1234}_{1234} = 2.885:2:35:977:696:\overbrace{\ldots}^\text{1227 digits}:0:1_{1234} = 2.717181235\ldots_{10}</math>
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