Pre-Calc Homework Solutions 369

# Pre-Calc Homework Solutions 369 - Section 9.1 38. The area...

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Unformatted text preview: Section 9.1 38. The area of each square is half of the area of the preceding square, so the total of all the areas is 4 4 n 0 369 44. Comparing 3 1 x3 with a 1 r , the first term is a 3 and 1 n 2 the common ratio is r Series: 3 3x 3 3x 6 x 3. ... 3x 3n ... 1, so the interval 1 1 2 8 m 2. 39. Total area n 1 2n 1 2 2 1 2 1 n 2 1 2 2n Interval: The series converges when x 3 of convergence is ( 1, 1). 45. Comparing 1 (x n 1 n 0 1 n 4 2 4 1 2 1 4) with a 1 r , the first term is a 4). ... ( 1)n(x 4 4)n 1 and the common ratio is r Series: 1 (x 4) (x 4)2 (x ... Interval: The series converges when x interval of convergence is (3, 5). 46. Comparing a 1 4 1 1 (x 1, so the 40. (a) S rS (a ar ar 2 ar 3 ... ar n 2 ar n 1) (ar ar 2 ar 3 ar 4 ... ar n 1 ar n) a ar n r: 1) with a 1 r , the first term is (x 1) 1 x. (b) Just factor and divide by 1 S S (1 rS r) S 41. Using the notation Sn a a ar n ar n 1 and the common ratio is r 4 1 (x 4 1 (x 4 Series: 1 4 1) 1)2 ... 1 ( 1)n(x 4 1)n ... a ar n 1 r Interval: The series converges when x interval of convergence is (0, 2). ar 2 ar 3 a ar . 1 r 1 n 1 1, so the a ar ... ar n 1, 47. Rewriting 1 2 x the formula from Exercise 40 is Sn If r n as 1 1 (x 1) and comparing with a 1 r , 1. 1, then lim r n a ar n 1 r n n 0 and so ar n a 1 r n 1 The first term is a Series: 1 (x 1) 1 and the common ratio is r (x 1) 2 x ... lim Sn lim . ... (x 1 1) n If r 1 or r 1, then r n has no finite limit as n , a ar n has no finite limit and 1 r n Interval: The series converges when x interval of convergence is (0, 2). Alternate solution: . Rewriting 1 2 1, so the so the expression diverges. If r ar n 1 1 1, then the nth partial sum is na, which goes to 1 1 3x 42. Comparing with a 1 r , the leading term is a 1 and the first is a 1 2 1 x 4 x 2 1 and the common ratio is r 2 1 2 x 8 as 1 1 x 2 1 and comparing with x . 2 a 1 r , the common ratio is r Series: 1 3x 9x 2 3x. ... ( 3x) n ... 3x 1, so the Series: ... 1 n x 2n 1 x 2 ... 1, so the interval Interval: The series converges when Interval: The series converges when interval of convergence is 43. Comparing x 1 2x 1 1 , . 3 3 a 1 r of convergence is ( 2, 2). 48. 1 eb e 2b e 3b ... with , the first term is a x and n 0 (e b)n 1 eb 1 the common ratio is r Series: x 2x 2 4x 3 2x. ... 2n 1 x n ... 1, so the interval 9 9 8 8 9 1 9e b eb b 9e b Interval: The series converges when 2x of convergence is 1 1 , . 2 2 ln 8 9 ln 8 ln 9 ...
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## This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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