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Unformatted text preview: SHORT CALCULUS Math 16C Sec 2 Spring 2008 Homework #6 Solutions Peter Malkin Section 10.2 Question 22 The infinite series n =0 4 ( 1 4 ) n is a geometric series. So, the sum of the series is a 1 r = 4 1 1 4 = 16 3 . Question 24 The series is a geometric series: 8 + 6 + 9 2 + 27 8 + ... = n =0 8 ( 3 4 ) n . So, the sum of the series is a 1 r = 8 1 3 4 = 32. Question 28 X n =0 [(0 . 7) n + (0 . 9) n ] = X n =0 (0 . 7) n + X n =0 (0 . 9) n . Now, n =0 (0 . 7) n is a geometric series where a = 1 and r = 0 . 7, which converges since  r  < 1, so X n =0 (0 . 7) n = a 1 r = 1 1 . 7 = 10 3 . Also, n =0 (0 . 9) n is a geometric series where a = 1 and r = 0 . 9, which converges since  r  < 1, so X n =0 (0 . 9) n = a 1 r = 1 1 . 9 = 10 . Therefore, X n =0 [(0 . 7) n + (0 . 9) n ] = X n =0 (0 . 7) n + X n =0 (0 . 9) n = 10 3 + 10 = 40 3 ....
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 Spring '11
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