The Binomial Theorem holds in any commutative ring

Assume R is a commutative ring with 1. Prove that the Binomial Theorem (a+b)^n = \sum_{k=0}^n {n \choose k} a^k b^{n-k} holds in R, where the binomial coefficient n \choose k is interpreted as the n \choose k-fold sum of 1.

We begin with some lemmas.

Recall that {n \choose k} = \frac{n!}{k!(n-k)!}, where n is a natural number and 0 \leq k < n.

Lemma: Let n be a natural number and 0 \leq k < n. Then the following hold.

  1. {n \choose 0} = {n \choose n} = 1.
  2. {{n+1} \choose {k+1}} = {n \choose k} + {n \choose {k+1}}.

Proof: The first facts follow directly from the definition of n \choose k. For the second, note the following.

\displaystyle\binom{n}{k+1} + \displaystyle\binom{n}{k}  =  \dfrac{n!}{(k+1)!(n-(k+1))!} + \dfrac{n!}{k!(n-k)!}
 =  \dfrac{n!(n-k)}{(k+1)!(n-k)!} + \dfrac{n!(k+1)}{(k+1)!(n-k)!}
 =  \dfrac{(n+1)!}{(k+1)!((n+1)-(k+1))!}
 =  \displaystyle\binom{n+1}{k+1} \square

We now prove the main result. Let a,b \in R. We proceed by induction on n; in a ring with 1, we use the convention that x^0 = 1. Then for the base case, we have (a+b)^0 = 1 = \binom{0}{0} a^0b^0.

For the inductive step, suppose (a+b)^n = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k}. Then we have the following.

(a+b)^{n+1}  =  (a+b)(a+b)^n
 =  (a+b)\left[ \displaystyle\sum_{k=0}^n \displaystyle\binom{n}{k} a^k b^{n-k} \right]
 =  \left[ \displaystyle\sum_{k=0}^n \displaystyle\binom{n}{k} a^{k+1} b^{n-k} \right] + \left[ \displaystyle\sum_{k=0}^n \displaystyle\binom{n}{k} a^kb^{n-k+1} \right]
 =  a^{n+1} + \left[ \displaystyle\sum_{k=0}^{n-1} \displaystyle\binom{n}{k} a^{k+1} b^{n-k} \right] + \left[ \displaystyle\sum_{k=1}^n \displaystyle\binom{n}{k} a^kb^{n-k+1} \right] + b^{n+1}
 =  a^{n+1} + \left[ \displaystyle\sum_{k=1}^n \displaystyle\binom{n}{k-1} a^{k} b^{n-k+1} \right] + \left[ \displaystyle\sum_{k=1}^n \displaystyle\binom{n}{k} a^kb^{n-k+1} \right] + b^{n+1}
 =  a^{n+1} + \left[ \displaystyle\sum_{k=1}^n \left[ \displaystyle\binom{n}{k-1} + \displaystyle\binom{n}{k} \right] a^kb^{(n+1)-k} \right] + b^{n+1}
 =  \displaystyle\binom{n+1}{n+1} a^{n+1}b^{(n+1)-(n+1)} + \left[ \displaystyle\sum_{k=1}^n \displaystyle\binom{n+1}{k} a^kb^{(n+1)-k} \right] + \displaystyle\binom{n+1}{0}a^0b^{(n+1)-0}
 =  \displaystyle\sum_{k=0}^{n+1} \displaystyle\binom{n+1}{k} a^k b^{(n+1)-k}.

Thus the result holds for all natural numbers n.

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