Proposition 1. Binomial Theorem.
Let $R$ be a non-trivial abelian ring and $n \in \N^+$. For any $a$, $b \in R \setdif \{0\}$,
\[(a + b)^n = \sum_{k = 0}^n \binom{n}{k} a^{n - k} b^k.\]
Proof. By induction:
- \[(a + b)^1 = a + b = \binom{1}{0} a^1 b^0 + \binom{1}{1} a^0 b^1.\]
-
Assume
\[(a + b)^n = \sum_{k = 0}^n \binom{n}{k} a^{n - k} b^k\]for some $n \in \N^+$. Then
\[\begin{align*} &\phantom{ {}={}} (a + b)^{n + 1} \\ &= (a + b)^n \cdot (a + b) \\ &= (a + b)^n \cdot a + (a + b)^n \cdot b \\ &= \sum_{k = 0}^n \binom{n}{k} a^{n + 1 - k} b^k +% \sum_{k = 0}^n \binom{n}{k} a^{n - k} b^{k + 1} \\ &= a^{n + 1} + \sum_{k = 1}^n \binom{n}{k} a^{n + 1 - k} b^k +% \sum_{k = 0}^{n - 1} \binom{n}{k} a^{n - k} b^{k + 1} + b^{n + 1} \\ &= a^{n + 1} + \sum_{k = 1}^n \binom{n}{k}a^{n + 1 - k} b^k +% \sum_{k = 1}^n \binom{n}{k - 1} a^{n + 1 - k} b^k + b^{n + 1} \\ &= a^{n + 1} + \sum_{k = 1}^n% \left[\binom{n}{k} + \binom{n}{k-1} \right] a^{n + 1 - k} b^k + b^{n + 1} \\ &= a^{n + 1} + \sum_{k = 1}^n \binom{n + 1}{k} a^{n + 1 - k} b^k + b^{n + 1} \\ &= \sum_{k = 0}^{n + 1} \binom{n + 1}{k} a^{n + 1 - k} b^k. \end{align*}\]
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