Editing Pascal's triangle (section)

=== Rows ===
* The sum of the elements of a single row is twice the sum of the row preceding it. For example, row&nbsp;0 (the topmost row) has a value of 1, row&nbsp;1 has a value of 2, row&nbsp;2 has a value of 4, and so forth. This is because every item in a row produces two items in the next row: one left and one right. The sum of the elements of row&nbsp;<math> n</math> equals to <math> 2^n</math>.
*Taking the product of the elements in each row, the sequence of products {{OEIS|id=A001142}} is related to the base of the natural logarithm, ''[[E (mathematical constant)|e]]''.<ref>{{citation
 | last = Brothers | first = H. J.
 | doi = 10.4169/math.mag.85.1.51
 | journal = [[Mathematics Magazine]]
 | pages = 51
 | title = Finding e in Pascal's triangle
 | volume = 85
 | year = 2012| issue = 1
 | s2cid = 218541210
 }}.</ref><ref>{{citation
 | last = Brothers | first = H. J.
 | doi =10.1017/S0025557200004204
 | journal = [[The Mathematical Gazette]]
 | pages = 145–148
 | title = Pascal's triangle: The hidden stor-''e''
 | volume = 96
 | year = 2012| issue = 535
 | s2cid = 233356674
 }}.</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.
 | doi = 10.5951/mathteacher.108.4.0246
 | journal = [[Mathematics Teacher]]
 | pages = 247
 | title = Nilakantha's Footprints in Pascal's Triangle
 | volume = 108
 | year = 2014}}</ref> <math display="block">\pi = 3 + \sum_{n = 1}^{\infty} (-1)^{n + 1} \frac{{2n + 1 \choose 1}}{{2n + 1 \choose 2}{2n + 2 \choose 2}}</math>
* Some of the numbers in Pascal's triangle correlate to numbers in [[Lozanić's triangle]].
* The sum of the squares of the elements of row&nbsp;{{mvar|n}} equals the middle element of row&nbsp;{{math|2''n''}}. For example, {{math|1=1<sup>2</sup> + 4<sup>2</sup> + 6<sup>2</sup> + 4<sup>2</sup> + 1<sup>2</sup> = 70}}. In general form, <math display="block">\sum_{k=0}^n {n \choose k}^2 = {2n \choose n}.</math>
* In any even row <math>n=2m</math>, the middle term minus the term two spots to the left equals a [[Catalan number]], specifically <math>C_{m-1} = \tbinom{2m}{m} - \tbinom{2m}{m-2}</math>. For example, in row&nbsp;4, which is 1, 4, 6, 4, 1, we get the 3rd Catalan number <math>C_3 = 6-1 = 5 </math>.
* In a row&nbsp;{{mvar|p}}, where {{mvar|p}} is a [[prime number]], all the terms in that row except the 1s are divisible by {{mvar|p}}. This can be proven easily, from the multiplicative formula <math>\tbinom pk = \tfrac{p!}{k!(p-k)!} </math>. Since the denominator <math>k!(p-k)!</math> can have no prime factors equal to {{mvar|p}}, so {{mvar|p}} remains in the numerator after integer division, making the entire entry a multiple of {{mvar|p}}.<!--
[[Image:Exp binomial grey.svg|thumb|upright=1.25|left|Binomial matrix as matrix exponential (illustration for 5×5 matrices). All the dots represent 0.]]  Comment: picture hidded since useless rescaling, also wrong paragraph for the pic -->
* ''Parity'': To count [[odd number|odd]] terms in row&nbsp;{{mvar|n}}, convert {{mvar|n}} to [[binary numeral system|binary]]. Let {{mvar|x}} be the number of 1s in the binary representation. Then the number of odd terms will be {{math|2<sup>''x''</sup>}}. These numbers are the values in [[Gould's sequence]].<ref>{{citation
 | last = Fine | first = N. J.
 | doi = 10.2307/2304500
 | journal = [[American Mathematical Monthly]]
 | mr = 0023257
 | pages = 589–592
 | title = Binomial coefficients modulo a prime
 | volume = 54
 | issue = 10
 | year = 1947| jstor = 2304500
 }}. See in particular Theorem 2, which gives a generalization of this fact for all prime moduli.</ref>
* Every entry in row 2<sup>''n''</sup>&nbsp;−&nbsp;1, ''n''&nbsp;≥&nbsp;0, is odd.<ref>{{citation
 | last = Hinz | first = Andreas M.
 | doi = 10.2307/2324061
 | issue = 6
 | journal = The American Mathematical Monthly
 | mr = 1166003
 | pages = 538–544
 | title = Pascal's triangle and the Tower of Hanoi
 | volume = 99
 | year = 1992| jstor = 2324061
 }}. Hinz attributes this observation to an 1891 book by [[Édouard Lucas]], ''Théorie des nombres'' (p.&nbsp;420).</ref>
*''Polarity'': When the elements of a row of Pascal's triangle are alternately added and subtracted together, the result is 0. For example, row 6 is 1,&nbsp;6,&nbsp;15,&nbsp;20,&nbsp;15,&nbsp;6,&nbsp;1, so the formula is 1&nbsp;−&nbsp;6&nbsp;+&nbsp;15&nbsp;−&nbsp;20&nbsp;+&nbsp;15&nbsp;−&nbsp;6&nbsp;+&nbsp;1&nbsp;=&nbsp;0.