Section 11.5 - Section 11.5 The Binomial Theorem(x a n The formulas for n = 1 2 3 and 4 are simple(x a)1 = x a(x a)2 = x2 2ax a2(x a)3 = x3 3ax2

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Unformatted text preview: Section 11.5 The Binomial Theorem (x + a) n The formulas for n = 1, 2, 3, and 4 are simple (x + a)1 = x + a (x + a)2 = x2 + 2ax + a2 (x + a)3 = x3 + 3ax2 + 3a2x + a3 (x + a)4 = x4 + 4ax3 + 6a2x2 + 4a3x + a4 But for n > 4 they get a bit more complicated. ( ) n j If j and n are integers with 0 j n, the symbol (nj), read as n take j at a time, is defined as (nj) = n! / [j!(n j)!] Some common formulas are (n0) = 1 (n1) = n (nn-1) = n (n n ) = 1 One that is not in the book is (nj) = (nn-j). The Pascal Triangle ( 00) (10) (11) (20) (21) (22) (30) (31) (32) (33) 1 1 1 1 3 2 3 1 1 1 The Pascal Triangle To Pascal triangle has 1's down the sides. To get any other entry you add the two nearest entries in the row above. This triangle in practice is not all that useful since for large values of n it is not practical to create the triangle. Binomial Theorem Let x and a be real numbers. For any positive integer n, we have (x + a)n = (n0) a0xn + (n1)a1xn-1 + ... + (nj)ajxn-j + ... + (nn)anx0 = (nj)ajxn-j Where the index j runs from 0 to n. Finding a Particular coefficient in a Binomial Expansion Based on the expansion of (x + a)n, the term containing xj is (nn-j)an-jxj Theorem If n and j are integers with 1 j n, then (nj-1) + (nj) = (n+1j). ...
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This note was uploaded on 08/26/2009 for the course MATH 125 taught by Professor Staff during the Fall '08 term at Southern Illinois University Edwardsville.

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