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Stitz-Zeager_College_Algebra_e-book

# Using proportionality arguments it stands to reason

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Unformatted text preview: pounding period, we earn interest on A2 which then grows to A2 (1 + i). We add our third payment at the end of the third compounding period to obtain A3 = A2 (1 + i) + P = P (1 + i) 1 + 1 1+i (1 + i) + P = P (1 + i)2 1 + 1 1 + 1 + i (1 + i)2 During the fourth compounding period, A3 grows to A3 (1+ i), and when we add the fourth payment, we factor out P (1 + i)3 to get A4 = P (1 + i)3 1 + 1 1 1 + + 2 1 + i (1 + i) (1 + i)3 This pattern continues so that at the end of the k th compounding, we get Ak = P (1 + i)k−1 1 + 4 1 1 1 + + ... + 1 + i (1 + i)2 (1 + i)k−1 This is the identity function on the natural numbers! There is an interesting anecdote which says that the famous mathematician Carl Friedrich Gauss was given this problem in primary school and devised a very clever solution. 6 The reader may wish to re-read the discussion on compound interest in Section 6.5 before proceeding. 5 568 Sequences and the Binomial Theorem The sum in the parentheses above is the sum of the ﬁrst k terms of a geometric sequence with 1 a = 1 and r = 1+i . Using Equation 9.2, we get 1 1− −k (1 + i)k = (1 + i) 1 − (1 + i) = 1 1 i 1− 1+i 1+ 1 1 1 + .....
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