Stitz-Zeager_College_Algebra_e-book

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Unformatted text preview: and n = 3, we get 1 2+ 10 3 3 = j =0 = 3 3−j 2 j 3 3−0 2 0 = 8+ 1 10 1 10 0 + j 3 3−1 2 1 12 6 1 + + 10 100 1000 = 8 + 1.2 + 0.06 + 0.001 = 9.261 1 10 1 + 3 3−2 2 2 1 10 2 + 3 3−3 2 3 1 10 3 588 Sequences and the Binomial Theorem 3. Identifying a = 2x, b = y and n = 5, the Binomial Theorem gives 5 5 (2x + y ) = j =0 5 (2x)5−j y j j Since we are concerned with only the term containing x3 , there is no need to expand the entire sum. The exponents on each term must add to 5 and if the exponent on x is 3, the exponent on y must be 2. Plucking out the term j = 2, we get 5 (2x)5−2 y 2 = 10(2x)3 y 2 = 80x3 y 2 2 We close this section with Pascal’s Triangle, named in honor of the mathematician Blaise Pascal. Pascal’s Triangle is obtained by arranging the binomial coefficients in the triangular fashion below. 0 0 1 0 2 0 3 0 4 0 1 1 2 1 3 1 4 1 2 2 3 2 4 2 . . . 3 3 4 3 4 4 Since n = 1 and n = 1 for all whole numbers n, we get that each row of Pascal’s Triangle 0 n begins and ends with 1. To generate the...
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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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