binomial theorem, proof of


Proposition.

Let a and b be commuting elements of some rig. Then

(a+b)n=∑k=0n(nk)⁢ak⁢bn-k,

where the (nk) are binomial coefficientsMathworldPlanetmath.

Proof.

Each term in the expansion of (a+b)n is obtained by making n decisions of whether to use a or b as a factor. Moreover, any sequenceMathworldPlanetmath of n such decisions yields a term in the expansion. So the expandsion of (a+b)n is precisely the sum of all the ab-words of length n, where each word appears exactly once.

Since a and b commute, we can reduce each term via rewrite rules of the form bn⁢a↦a⁢bn to a term in which the a factors precede all the b factors. This produces a term of the form ak⁢bn-k for some k, where we use the expressions a0⁢bn and an⁢b0 to denote bn and an respectively. For example, reducing the word b⁢a⁢b⁢a⁢b2⁢a⁢b⁢a yields a4⁢b5, via the following reduction.

b⁢a⁢b⁢a⁢b2⁢a⁢b⁢a↦a⁢b⁢a⁢b⁢a⁢b2⁢a⁢b↦a2⁢b⁢a⁢b⁢a⁢b3↦a3⁢b⁢a⁢b4↦a4⁢b5.

After performing this rewriting process, we collect like terms. Let us illustrate this with the case n = 3.

(a+b)3 =a⁢a⁢a+a⁢a⁢b+a⁢b⁢a+a⁢b⁢b+b⁢a⁢a+b⁢a⁢b+b⁢b⁢a+b⁢b⁢b
=a3+a2⁢b+a2⁢b+a⁢b2+a2⁢b+a⁢b2+a⁢b2+b3
=a3+3⁢a2⁢b+3⁢a⁢b2+b3.

To determine the coefficient of a reduced term, it suffices to determine how many ab-words have that reduction. Since reducing a term only changes the positions of as and bs and not their number, all the ab-words where k of the letters are bs and n-k are as, for 0≤k≤n, have the same normalization. But there are exactly (nk) such ab-words, since there are (nk) ways to select k positions out of n to place as in an ab-word of length n. This shows that the coefficient of the an term is (n0)=1, the coefficient of the bn term is (nn)=1, and that the coefficient of the ak⁢bn-k term is (nk). ∎

Title binomial theoremMathworldPlanetmath, proof of
Canonical name BinomialTheoremProofOf
Date of creation 2013-03-22 15:03:54
Last modified on 2013-03-22 15:03:54
Owner mps (409)
Last modified by mps (409)
Numerical id 15
Author mps (409)
Entry type Proof
Classification msc 11B65