Stitz-Zeager_College_Algebra_e-book

# In dening the cosine and sine functions we assigned

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Unformatted text preview: numbers in the middle of the rows (from the third row onwards), we take advantage of the additive relationship expressed in Theorem 9.3. For instance, 1 1 2 2 2 3 0 + 1 = 1 , 0 + 1 = 1 and so forth. This relationship is indicated by the arrows in the array above. With these two facts in hand, we can quickly generate Pascal’s Triangle. We start with the ﬁrst two rows, 1 and 1 1. From that point on, each successive row begins and ends with 1 and the middle numbers are generated using Theorem 9.3. Below we attempt to demonstrate this building process to generate the ﬁrst ﬁve rows of Pascal’s Triangle. 9.4 The Binomial Theorem 589 1 1 1 1 1 1+1 1 −− −→ −−− 1 1 2 1 1 1 1 1+2 1 2+1 1 −− −→ −−− 1 1 1 1 3 1 1 3 1+3 1 1 3 3+3 1 3 1 1 1 2 1 2 1 1 1 1 1 2 1 3+1 1 −− −→ −−− 1 1 1 1 1 2 3 4 1 3 6 1 4 1 To see how we can use Pascal’s Triangle to expedite the Binomial Theorem, suppose we wish to expand (3x − y )4 . The coeﬃcients we need are 4 for j = 0, 1, 2, 3, 4 and are the numbers which j form the ﬁfth row of Pascal’s Triangle. Since we know that the exponent of 3x in the ﬁrst term is 4 and then decreases by one as we go from left to right while the exponent of −y starts at 0 in the ﬁrst term and then...
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