Pascal's triangle: Difference between revisions

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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'' +1 into ''k'' +1 ordered partitions form Pascal's triangle.]]

=== 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>{{citation

| last = Foster | first = T.

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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 [[Negative base|holds]] for <math>a =\bmod 2c</math>, with <math>a</math> congruent to <math>\{c - 1, -(c + 1)\} \;\mathrm{mod}\; 2c</math>, and with odd values of <math>n</math> [[Negative number#Multiplication|yielding]] negative row products.<ref>{{cite book

| display-authors = etal

| last = Hilton | first = P.