MATH 152 Midterm: MATH 152 TAMU 2009a Exam 3b Solutions
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Solutions-form b: d: recall an = lim n sn, where sn is the sequence. We are given that sn = 4 + ln(n + 1) ln(2n), thus. Pn=1 n (cid:18)4 + ln (4 + ln(n + 1) ln(2n)) n + 1. 2: e: collect all variables on the left hand side of the equation and complete the squares: x2 + 2x + y2 4y + z2 6z = 0. Complete the three squares: x2 + 2x + 1 + y2 4y + 4 + z2 6z + 9 = 14 (x + 1)2 + (y 2)2 + (z 3)2 = 14. Thus the center of the sphere is ( 1, 2, 3) and the radius is 14: a: t3(x) = f (i)(1) i! Pi=0 (x 1)i f (1) f (1) (x 1)2 + (x 1)3. 1 (x 1)2 + (x 1)3. 5: c: the series is a convergent geometric series with r = false.
