Editing Pascal's triangle (section)

=== Calculating a row or diagonal by itself ===
There are simple algorithms to compute all the elements in a row or diagonal without computing other elements or factorials.

To compute row <math>n</math> with the elements <math>\tbinom{n}{0}, \tbinom{n}{1}, \ldots, \tbinom{n}{n}</math>, begin with <math>\tbinom{n}{0}=1</math>. For each subsequent element, the value is determined by multiplying the previous value by a fraction with slowly changing numerator and denominator:

:<math> {n\choose k}= {n\choose k-1}\times \frac{n+1-k}{k}.</math>

For example, to calculate row 5, the fractions are&nbsp; <math>\tfrac{5}{1}</math>,&nbsp; <math>\tfrac{4}{2}</math>,&nbsp; <math>\tfrac{3}{3}</math>,&nbsp; <math>\tfrac{2}{4}</math> and <math>\tfrac{1}{5}</math>, and hence the elements are&nbsp; <math>\tbinom{5}{0}=1</math>, &nbsp; <math>\tbinom{5}{1}=1\times\tfrac{5}{1}=5</math>, &nbsp; <math>\tbinom{5}{2}=5\times\tfrac{4}{2}=10</math>, etc. (The remaining elements are most easily obtained by symmetry.)

To compute the diagonal containing the elements <math>\tbinom{n}{0}, \tbinom{n+1}{1}, \tbinom{n+2}{2},\ldots,</math> begin again with <math>\tbinom{n}{0} = 1</math> and obtain subsequent elements by multiplication by certain fractions:

:<math> {n+k\choose k}= {n+k-1\choose k-1}\times \frac{n+k}{k}.</math>

For example, to calculate the diagonal beginning at <math>\tbinom{5}{0}</math>, the fractions are&nbsp; <math>\tfrac{6}{1}, \tfrac{7}{2}, \tfrac{8}{3}, \ldots</math>, and the elements are <math>\tbinom{5}{0}=1, \tbinom{6}{1}=1 \times \tfrac{6}{1}=6, \tbinom{7}{2}=6\times\tfrac{7}{2}=21</math>, etc. By symmetry, these elements are equal to <math>\tbinom{5}{5}, \tbinom{6}{5}, \tbinom{7}{5}</math>, etc.

[[File:Fibonacci in Pascal triangle.png|thumb|[[Fibonacci sequence]] in Pascal's triangle]]