Let $R$ be a non-trivial abelian ring and $n\in\N^+$. For any $a$, $b\in R^*$,
\[(a+b)^n = \sum_{k=0}^n\binom{n}{k}a^{n-k}b^k.\]
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
\[\begin{align*} &\phantom{ {}={}} (a+b)^{n+1} \\ &= (a+b)^n\cdot(a+b) \\ &= (a+b)^n\cdot a+(a+b)^n\cdot b \\ &= \sum_{k=0}^n\binom{n}{k}a^{n+1-k}b^k+\sum_{k=0}^n\binom{n}{k}a^{n-k}b^{k+1} \\ &= a^{n+1}+\sum_{k=1}^n\binom{n}{k}a^{n+1-k}b^k+\sum_{k=0}^{n-1}\binom{n}{k}a^{n-k}b^{k+1}+b^{n+1} \\ &= a^{n+1}+\sum_{k=1}^n\binom{n}{k}a^{n+1-k}b^k+\sum_{k=1}^{n}\binom{n}{k-1}a^{n+1-k}b^k+b^{n+1} \\ &= a^{n+1}+\sum_{k=1}^n\left[\binom{n}{k}+\binom{n}{k-1}\right]a^{n+1-k}b^k+b^{n+1} \\ &= a^{n+1}+\sum_{k=1}^n\binom{n+1}{k}a^{n+1-k}b^k+b^{n+1} \\ &= \sum_{k=0}^{n+1}\binom{n+1}{k}a^{n+1-k}b^k. \end{align*}\]