Editing Binomial theorem (section)

=== Inductive proof ===
[[mathematical induction|Induction]] yields another proof of the binomial theorem. When {{math|1=''n'' = 0}}, both sides equal {{math|1}}, since {{math|1=''x''<sup>0</sup> = 1}} and <math>\tbinom{0}{0}=1.</math>  Now suppose that the equality holds for a given {{mvar|n}}; we will prove it for {{math|1=''n'' + 1}}.  For {{math|1=''j'', ''k'' ≥ 0}}, let {{math|1=[''f''(''x'', ''y'')]<sub>''j'',''k''</sub>}} denote the coefficient of {{math|1=''x''<sup>''j''</sup>''y''<sup>''k''</sup>}} in the polynomial {{math|1=''f''(''x'', ''y'')}}.  By the inductive hypothesis, {{math|1=(''x'' + ''y'')<sup>''n''</sup>}} is a polynomial in {{mvar|x}} and {{mvar|y}} such that {{math|1=[(''x'' + ''y'')<sup>''n''</sup>]<sub>''j'',''k''</sub>}} is <math>\tbinom{n}{k}</math> if {{math|1=''j'' + ''k'' = ''n''}}, and {{mvar|0}} otherwise.  The identity
<math display="block"> (x+y)^{n+1} = x(x+y)^n + y(x+y)^n</math>
shows that {{math|1=(''x'' + ''y'')<sup>''n''+1</sup>}} is also a polynomial in {{mvar|x}} and {{mvar|y}}, and
<math display="block"> [(x+y)^{n+1}]_{j,k} = [(x+y)^n]_{j-1,k} + [(x+y)^n]_{j,k-1},</math>
since if {{math|1=''j'' + ''k'' = ''n'' + 1}}, then {{math|1=(''j'' − 1) + ''k'' = ''n''}} and {{math|1=''j'' + (''k'' − 1) = ''n''}}. Now, the right hand side is
<math display="block"> \binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k},</math>
by [[Pascal's identity]].<ref>[http://proofs.wiki/Binomial_theorem Binomial theorem] – inductive proofs {{webarchive |url=https://web.archive.org/web/20150224130932/http://proofs.wiki/Binomial_theorem |date=February 24, 2015 }}</ref> On the other hand, if {{math|1=''j'' + ''k'' ≠ ''n'' + 1}}, then {{math|1=(''j'' – 1) + ''k'' ≠ ''n''}} and {{math|1=''j'' + (''k'' – 1) ≠ ''n''}}, so we get {{math|1=0 + 0 = 0}}. Thus
<math display="block">(x+y)^{n+1} = \sum_{k=0}^{n+1} \binom{n+1}{k} x^{n+1-k} y^k,</math>
which is the inductive hypothesis with {{math|1=''n'' + 1}} substituted for {{mvar|n}} and so completes the inductive step.