Binomial theorem for a commutative ring
1. Satz
Sei \(R\) ein kommutativer Ring und seien \(x,y \in R\) sowie \(n \in \mathbb{N}\). Dann gilt:
\begin{align*} (x + y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k}y^{k} \end{align*}2. Beweis
2.1. Induktionsanfang
\begin{align*}
(x + y)^0 = 1 = \sum_{k=0}^0 \binom{0}{0} x^0 y^0 = 1
\end{align*}
2.2. Induktionsschritt
\begin{align*}
(x + y)^{n+1} =& (x+y) (x+y)^n \\
=& x(x+y)^n + y(x+y)^n \\
=& x \cdot \sum_{k=0}^n \binom{n}{k} x^{n-k}y^k + y \cdot \sum_{k=0}^n \binom{n}{k} x^{n-k}y^{k} \\
=& \sum_{k=0}^n \binom{n}{k} x^{n+1-k}y^k + \sum_{k=0}^n \binom{n}{k} x^{n-k}y^{k+1}
\end{align*}
Anwenden des Lemmas Binomialkoeffizient - k > n
\begin{align*} \binom{n}{n+1} =& 0 \\ \Rightarrow \sum_{k=0}^{n} \binom{n}{k} a_k =& \sum_{k=0}^{n} \binom{n}{k} a_k + 0 \\ =& \sum_{k=0}^{n} \binom{n}{k} a_k + \binom{n}{n+1} a_{n+1} \\ =& \sum_{k=0}^{n+1} \binom{n}{k} a_k \\ \Rightarrow \sum_{k=0}^n \binom{n}{k} x^{n+1-k}y^k =& \sum_{k=0}^{n+1} \binom{n}{k} x^{n+1-k}y^{k} \end{align*} \begin{align*} \sum_{k=0}^n \binom{n}{k} x^{n-k}y^{k+1} \end{align*}2.2.1. y
\begin{align*}
\sum_{k=0}^n a_{k+1} =& \sum_{k=1}^{n+1} a_k \\
\Rightarrow \sum_{k=0}^n \binom{n}{k} x^{n-k}y^{k+1} =& \sum_{k=1}^{n+1}\binom{n}{k-1} x^{n-(k-1)}y^k \\
=& \sum_{k=1}^{n+1}\binom{n}{k-1} x^{n+1-k}y^k \\
\end{align*}
\begin{align*}
\binom{n}{-1} =& 0 \\
\Rightarrow \sum_{k=1}^{n+1}\binom{n}{k-1} x^{n+1-k}y^k =& \sum_{k=1}^{n+1}\binom{n}{k-1} x^{n+1-k}y^k + \binom{n}{-1} x^{n+1}y^{0}\\
=& \sum_{k=0}^{n+1}\binom{n}{k-1} x^{n+1-k}y^k
\end{align*}
2.2.2. Zusammenaddieren
\begin{align*}
&\sum_{k=0}^n \binom{n}{k} x^{n+1-k}y^k + \sum_{k=0}^n \binom{n}{k} x^{n-k}y^{k+1} \\
=& \sum_{k=0}^{n+1} \binom{n}{k} x^{n+1-k}y^{k} + \sum_{k=0}^{n+1} \binom{n}{k-1}x^{n+1-k}y^k \\
=& \sum_{k=0}^{n+1} \left( \binom{n}{k} x^{n+1-k}y^k + \binom{n}{k-1} x^{n+1-k}y^k \right) \\
=& \sum_{k=0}^{n+1} \left( \binom{n+1}{k} x^{n+1-k}y^k\right)
\end{align*}