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Induction step of the proof of the binomial theorem Assume for some arbitrary natural number m, that for arbitrary a and b: m m m m (a b) m a m a m1b1 a m2b2 ... b m 0 1 2 m Consider case m+1. We need to show the binomial equation works for this case. Expand the left side of (a b) m1 by the distributive law: (a b) m1 (a b)(a b) m m m m m a a m a m1b1 a m2 b 2 ... b m 1 2 m 0 m m m m b a m a m1b1 a m2 b 2 ... b m 1 2 m 0 by using the induction hypothesis twice. Now multiply through each term of the first row by a and each term of the second row by b: m m1 m m 1 m m1 2 m m a a b a b ... ab 0 1 2 m m m 1 m m1 2 m m2 3 m m1 a b a b a b ... b 0 1 2 m Then carefully combine similar terms (same powers of a and b). Notice each term (after the first) is similar to the term below and to the left. The first and last terms have nothing similar. So: m m m m m1 m m1 m m m 1 m m m1 2 a a b a b ... ab b 1 0 2 1 m m 1 0 m (Almost done.) Now notice the first and last binomial coefficients are 1. We could shift the m’s to m+1 and they are still 1. And then – notice Pascal’s rule allows us to rewrite each of the sums of 2 binomial coefficients above. So we get exactly what we want-the right side: m 1 m1 m 1 m 1 m 1 m1 2 m 1 m m 1 m1 a a b a b ... ab b 0 1 2 m m 1