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Math 618 Answer to Problem 2.3.11 Autumn 2005 Algebra/Calculus Reminder Let Sn = (Ia)n| = v + 2v2 + 3v3 + ··· + nvn. We have the formula: Sn = (Ia)n| = v + 2v2 + 3v3 + ··· + nvn = ¨an| − nv n i . Using ¨an| = 1 + v + v2 + ··· + vn−1 = 1 − v n 1 − v in the above formula, we have Sn = (Ia)n| = 1−vn 1−v − nv n i . If |v| < 1, then limn→∞vn = 0 and limn→∞nvn = 0 (why?). Let n → 0 in the formula of Sn above, we get the increasing perprtuity: S∞ = v + 2v2 + 3v3 + 4v4 + ··· = (Ia)∞| = parenleftbigg 1 1 − v parenrightbigg /i = 1i + 1i2. (1) We also have a∞| = v + v2 + v3 + ··· + vk + ··· = 1i, (2) and ¨a∞| = 1 + v + v2 + ··· + vk + ··· = 11 − v = 1 + ii . (3) Solution Sandy’s perpetuity can be explained by the following diagram (v = 11+i): . . . . . . . . .0 1 2 3 k 90 + 10 90 + 20 90 + 30 . . . 90 + 10k . . . (90 + 10)v (90 + 20)v2 (90 + 30)v3 . . . (90 + 10k)vk . . . a27 a27 a27 a27 a27 a27 The sum of these present values is (using (1) and (2)): (90 + 10)v + (90 + 20)v2 + (90 + 30)v3 + ··· + (90 + 10k)vk + ··· = (90v + 90v2 + 90v3 + ··· + 90vk + ··· ) + (10v + 20v2 + 30v3 + ··· + 10kvk + ··· ) = 90a∞| + 10(Ia)∞| = 901i + 10 parenleftbigg1 i + 1 i2 parenrightbigg = 100i + 10i2 (4) Similarly, Danny’s perpetuity-due can be explained by the following diagram: . . . . . . . . .0 1 2 k 180 180 . . . 180 . . . 180 180v 180v2 . . . 180vk . . . a27 a27 a27 a27 a27 The sum of these present values is: 180 + 180v + 180v2 + ··· + 180vk + ··· = 180¨a∞| = 1801 + ii . (5) Since the two present values in (4) and (5) are equal, we have 100 i + 10 i2 = 180 1 + i i . Simplifying the last equation, we get a quadratic equation 18i2 +8i−1 = 0. Solving this equation (note that i > 0), we get i = 0.10171955. Answer: i = 0.1017

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