section_heading: Dokaz indukcijom
paragraph_text: Prema pretpostavci, sada pišemo <math>(a + b)^{n + 1} = (a + b)\sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k </math>, što daje <math>\sum_{k=0}^{n} \binom{n}{k} a^{n+1-k} b^k + \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k + 1} </math>.