On part comme précédemment :
\[
(a+b)^5 = (a+b)\,(a+b)^4
\]
mais en écrivant :
\[
(a+b)^4 = \lambda_{4,0}\,a^4+\lambda_{4,1}\,a^3\,b+\lambda_{4,2}\,a^2\,b^2
+\lambda_{4,3}\,a\,b^3+\lambda_{4,4}\,b^4.
\]
En développant, on obtient :
\begin{align*}
(a+b)^5 &= (a+b)\,(a+b)^4\\ \\
&= (a+b)\,\big(\lambda_{4,0}\,a^4+\lambda_{4,1}\,a^3\,b+\lambda_{4,2}\,a^2\,b^2
+\lambda_{4,3}\,a\,b^3+\lambda_{4,4}\,b^4\big)\\ \\
&=\lambda_{4,0}\,a^5+\lambda_{4,1}\,a^4\,b+\lambda_{4,2}\,a^3\,b^2
+\lambda_{4,3}\,a^2\,b^3+\lambda_{4,4}\, a\,b^4\\ \\
&\phantom{=\lambda_{4,0}\,a^5\;}+\lambda_{4,0}\,a^4\,b+\lambda_{4,1}\,a^3\,b^2
+\lambda_{4,2}\,a^2\,b^3+\lambda_{4,3}\,a\,b^4+\lambda_{4,4}\,b^5.
\end{align*}
Si l'on pose alors :
\[
\lambda_{5,0} = \lambda_{4,0} \et[] \lambda_{5,5} = \lambda_{4,4}
\]
ainsi que :
\[
\forall{k\in\ceo1,4\cef}\quad \lambda_{5,k} = \lambda_{4,k}+ \lambda_{4,k-1}
\]
on a :
\[
(a+b)^5 = \lambda_{5,0}\,a^5+\lambda_{5,1}\,a^4\,b+\lambda_{5,2}\,a^3\,b^2
+\lambda_{5,3}\,a^2\,b^3+\lambda_{5,4}\,a\,b^4+\lambda_{5,5}\,b^4,
\]
ou encore :
\[
(a+b)^5 = \sum_{k=0}^n\,\lambda_{5,k}\,a^{n-k}\,b^k.
\]