Stitz-Zeager_College_Algebra_e-book

Rewriting the identity using this convention results

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Unformatted text preview: instances of the formula, namely (a + b)1 = a + b (a + b)2 = a2 + 2ab + b2 (a + b)3 = a3 + 3a2 b + 3ab2 + b3 If we wanted the expansion for (a + b)4 we would write (a + b)4 = (a + b)(a + b)3 and use the formula that we have for (a+b)3 to get (a+b)4 = (a+b) a3 + 3a2 b + 3ab2 + b3 = a4 +4a3 b+6a2 b2 +4ab3 +b4 . Generalizing this a bit, we see that if we have a formula for (a + b)k , we can obtain a formula for (a + b)k+1 by rewriting the latter as (a + b)k+1 = (a + b)(a + b)k . Clearly this means Mathematical Induction plays a major role in the proof of the Binomial Theorem.1 Before we can state the theorem we need to revisit the sequence of factorials which were introduced in Example 9.1.1 number 6 in Section 9.1. Definition 9.4. Factorials: For a whole number n, n factorial, denoted n!, is the term fn of the sequence f0 = 1, fn = n · fn−1 , n ≥ 1. Recall this means 0! = 1 and n! = n(n − 1)! for n ≥ 1. Using the recursive definition, we get: 1! = 1 · 0! = 1 · 1 = 1, 2! = 2 ...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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