Geometric Series for Catching Fish Each DayDate: 05/24/2007 at 12:45:34 From: Amanda Subject: Percents and fishing The first day she went fishing, she caught exactly 31416 pounds of fish. The next day she caught 40% of what she caught the day before, which was still a lot of fish. If she continued fishing every day for twenty years and each day caught 40% of what she caught the day before, how many pounds of fish, total, would she catch? Please round to the nearest pound. Date: 05/24/2007 at 13:38:11 From: Doctor Ian Subject: Re: Percents and fishing Hi Amanda, Let's try something simpler, but similar. Suppose I catch 1000 pounds of fish, then 1/3 of that the next day, then 1/3 of that the next day, and so on for 5 days. How much fish is that? The first day, I catch 1000 lbs. The second day, I catch (1/3)*1000 pounds. The third day, I catch (1/3)*(1/3)*1000 and so on. After 5 days, my total will be 1000 * 1 1000 * (1/3) 1000 * (1/3)^2 1000 * (1/3)^3 + 1000 * (1/3)^4 ---------------- ? But that will be 1000 * (1 + (1/3) + (1/3)^2 + (1/3)^3 + (1/3)^4) and this part, (1 + (1/3) + (1/3)^2 + (1/3)^3 + (1/3)^4) is the sum of a geometric series. We can find such a sum using the formula described here: Geometric Sequences and Series http://mathforum.org/library/drmath/view/56960.html that is, 1(1 - (1/3)^5) -------------- 1 - 1/3 1 - 1/243 = --------- 2/3 242/243 = ------- 2/3 242 3 = --- * - 243 2 = 1.49 (approximately) This case is small enough to check by hand: (1 + (1/3) + (1/3)^2 + (1/3)^3 + (1/3)^4) = 1 + 1/3 + 1/9 + 1/27 + 1/81 = 1 + 27/81 + 9/81 + 3/81 + 1/81 = 1 + 40/81 = 1.49 (approximately) So we can have some confidence that this is the right formula! We'd multiply 1.49 by 1000, to get 1490 pounds of fish. Does that make sense? If so, note that your problem is exactly the same, except you have 1) a different starting amount (31416 instead of 1000), 2) a different ratio (2/5, instead of 1/3), and 3) a different number of days (the number of days in twenty years, instead of 5 days). Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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