EECS 1019 Study Guide - Midterm Guide: Junkers J 1, Mathematical Induction, Mater Lectionis
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EECS 1019 Full Course Notes
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Math/eecs 1019 third test (version 1) fall 2014. Solutions: (3 points) find the sum 1*2+2*3+3*4+. 99(cid:88) i=1 i(i + 1) (i2 + i) = 33 50 199 + 99 50. = 333300: (3 points) a y starts at the origin and goes 1 unit up, 1/2 unit right, 1/4 unit down, 1/8 unit left, 1/16 unit up, etc. , ad in nitum. Solution: let us compute the x, y coordinates separately. The nal x coordinate is given by the in nite sum. This is a geometric series with a = 1. 1 r = 1/2 geometric series we get x = a. 2 and r = 1: so using the formula for the in nite. Similarly, the nal y coordinate is given by the in nite sum. This is a geometric series with a = 1 and r = 1 geometric series we get y = a.