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Unformatted text preview: ACF329 Interest Theory Fall 2010 University of Texas at Austin HW 1  Solutions Instructor: Milica ˇ Cudina 1.3.2 A 2000 (5) = 2000 a (5) = 2000(1 + 0 . 04 · 5) = 2000 · 1 . 2 = 2400 . 1.3.4 From a (0) = 1, we conclude that γ = 1. Next, the given conditions are 100( α · 4 2 + β · 4 + 1) = 152 , 200( α · 2 2 + β · 2 + 1) = 240 . Tidying up the expressions above, we get the following system of two equations with two unknowns α and β : 4 α + β = 0 . 13 , 2 α + β = 0 . 10 . So, α = 0 . 015 and β = 0 . 07 and a ( t ) = 0 . 015 t 2 + 0 . 07 t + 1 for t ≥ 0. Let B denote the amount in the account at time 8 if 1 , 600 was deposited at time 6. Then, B = 1600 · a (8) a (6) = 1600 · . 015 · 64+0 . 07 · 8+1 . 015 · 36+0 . 07 · 6+1 = 1600 · 2 . 52 1 . 96 ≈ 2057 . 14 . 1.3.6 According to the information from the problem, the amount of interest accrued from time 0 to time n is I n = 2 + 2 2 + ··· + 2 n 1 + 2 n . Using the following formula for the sum of the first n terms of a geometric sequence: 1 + Q + Q 2 + ··· + Q n 1 = 1 Q n 1 Q , we get I n = 2(1 + 2 + ··· + 2 n 1 ) = 2 · 1 2 n 1...
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This note was uploaded on 12/05/2011 for the course M 341 taught by Professor Hietmann during the Spring '08 term at University of Texas.
 Spring '08
 HIETMANN

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