Editing Binomial coefficient (section)

== Pascal's triangle ==
[[Image:Pascal's triangle - 1000th row.png|150px|right|thumb|1000th row of Pascal's triangle, arranged vertically, with grey-scale representations of decimal digits of the coefficients, right-aligned. The left boundary of the image corresponds roughly to the graph of the logarithm of the binomial coefficients, and illustrates that they form a [[logarithmically concave sequence|log-concave sequence]].]]

{{Main|Pascal's triangle|Pascal's rule}}

[[Pascal's rule]] is the important [[recurrence relation]]
{{NumBlk2|:|<math> {n \choose k} + {n \choose k+1} = {n+1 \choose k+1},</math>|3}}
which can be used to prove by [[mathematical induction]] that <math> \tbinom n k</math> is a natural number for all integer ''n'' ≥ 0 and all integer ''k'', a fact that is not immediately obvious from [[#Definition_and_interpretations|formula (1)]]. To the left and right of Pascal's triangle, the entries (shown as blanks) are all zero.

Pascal's rule also gives rise to [[Pascal's triangle]]:
{|
|-
|0: || || || || || || || || ||1|| || || || || || || ||
|-
|1: || || || || || || || ||1|| ||1|| || || || || || ||
|-
|2: || || || || || || ||1|| ||2|| ||1|| || || || || ||
|-
|3: || || || || || ||1|| ||3|| ||3|| ||1|| || || || ||
|-
|4: || || || || ||1|| ||4|| ||6|| ||4|| ||1|| || || ||
|-
|5: || || || ||1|| ||5|| ||10|| ||10|| ||5|| ||1|| || ||
|-
|6: || || ||1|| ||6|| ||15|| ||20|| ||15|| ||6|| ||1|| ||
|-
|7: || ||1&nbsp;|| ||7&nbsp;|| ||21|| ||35|| ||35|| ||21|| ||7&nbsp;|| ||1&nbsp;||
|-
|8: ||1&nbsp;|| ||8&nbsp;|| ||28|| ||56|| ||70|| ||56|| ||28|| ||8&nbsp;|| ||1&nbsp;
|} <!--There is a wider cell made with &nbsp; in 1-digit columns, so triangle becomes more graphically symmetrical -->

Row number {{mvar|n}} contains the numbers <math>\tbinom{n}{k}</math> for {{math|1=''k'' = 0, …, ''n''}}. It is constructed by first placing 1s in the outermost positions, and then filling each inner position with the sum of the two numbers directly above. This method allows the quick calculation of binomial coefficients without the need for fractions or multiplications. For instance, by looking at row number 5 of the triangle, one can quickly read off that
:<math>(x + y)^5 = \underline{1}x^5 + \underline{5}x^4y + \underline{10}x^3y^2 + \underline{10}x^2y^3 + \underline{5}xy^4 + \underline{1}y^5.</math>