The reference angle here is not so the reference

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Unformatted text preview: and we note that P (1) is true since 1 ? (a + b)1 = j =0 ? a+b = 1 1−j j ab j 1 1−0 0 1 1−1 1 a b+ ab 0 1 a+b = a+b Now we assume that P (k ) is true. That is, we assume that we can expand (a + b)k using the formula given in Theorem 9.4 and attempt to show that P (k + 1) is true. (a + b)k+1 = (a + b)(a + b)k k = (a + b) j =0 k =a j =0 k = j =0 k k−j j ab j k k−j j a b +b j k k+1−j j a b+ j k j =0 k j =0 k k−j j ab j k k−j j +1 ab j Our goal is to combine as many of the terms as possible within the two summations. As the counter j in the first summation runs from 0 through k , we get terms involving ak+1 , ak b, ak−1 b2 , . . . , abk . In the second summation, we get terms involving ak b, ak−1 b2 , . . . , abk , bk+1 . In other words, apart from the first term in the first summation and the last term in the second summation, we have 4 and a fair amount of tenacity and attention to detail. 586 Sequences and the Binomial Theorem terms common to both summations. Our next...
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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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