142p23s11solution[1]

# 9 more feet at the second return 21599 at the third

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Unformatted text preview: + a 6 + ... + a 40 ) n =1 ( = 3a 2 1 + ( a 21 Or, ∑ 3a n =1 20 2n = 3a 2 21 23 ) + (a ) + (a ) ∑ (a ) n =0 22 2n 2 = 3a ⋅ + ... + ( a 1 − ( a2 ) 1 − a2 21 . 2 20 ) ) = 3a ⋅ 2 1 − ( a2 ) 1− a2 21 MTH 142 Exam 3 Spr 2011 Practice Problem Solutions 1 No calculators will be permitted at the exam. 3.1 A ping-pong ball is launched straight up, rises to a height of 15 feet, then falls back to the launch point and bounces straight up again. It continues to bounce, each time reaching a height 90% of the height reached on the previous bounce. Find the total distance that the ball travels. The ball has gone 30 ft at the ﬁrst return to the launch point, then 2(15)(.9) more feet at the second return, 2(15)(.9)(.9) at the third return, etc. The total distance is then 30 = 300feet. 30 + 30(.9) + 30(.9)2 + 30(.9)3 + · · · = 1 − .9 3.2 Use the integral test to determine the convergence of the following series: b) ∞ 1 n=1 a) n 3/2 ∞ n n n=1 e a) Series converges if and only the improper integral ∞ − 3/2 dx = 2 < ∞ so series converges. 1x ∞ 1 x−3/2 dx converges. b) Series converges if and on...
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## This note was uploaded on 02/04/2014 for the course MAT 132 taught by Professor Poole during the Fall '08 term at SUNY Stony Brook.

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