The total amount spent is an infinite geometric series with a Nk and r k Since

# The total amount spent is an infinite geometric

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The total amount spent is an infinite geometric series with a = Nk and r = k . Since 0 < k < 1 , the series converges. We use the formula for the sum to see Total amount spent = Nk 1 - k = N k 1 - k . (b) We substitute k = 0 . 85 into the formula from part (a): Total amount spent = N k 1 - k = N 0 . 85 1 - 0 . 85 = 5 . 667 N. The total additional spending is more than 5 times the size of the original tax rebate. Solutions for Section 11.3 1. In 2003 , the total quantity of oil consumed was 28 . 5 billion barrels; each subsequent year, the quantity is multiplied by 1 . 021 . Thus, in 2004 , we have 28 . 5(1 . 021) = 29 . 1 billion barrels; in 2004 we have 28 . 5(1 . 021) 2 = 29 . 7 billion barrels; and so on. Year 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 Oil 28 . 5 29 . 1 29 . 7 30 . 3 31 . 0 31 . 6 32 . 3 33 . 0 33 . 7 34 . 4 In 2012 , the total quantity of oil consumed is 28 . 5(1 . 021) 9 . Thus, we sum the geometric series: Total consumption over decade = 28 . 5 + 28 . 5(1 . 021) + 28 . 5(1 . 021) 2 + · · · + 28 . 5(1 . 021) 9 = 28 . 5(1 - (1 . 021) 10 ) 1 - 1 . 021 = 313 . 498 billion barrels. 2. If consumption changes by a factor of r each year (so we use r = 0 . 995 in part (a) and r = 1 . 025 in part (b)), then consumption in 2004 is predicted to be 27 r billion barrels, in 2005 it is predicted to be 27 r 2 billion barrels, in 2006 it is predicted to be 27 r 3 billion barrels, and so on. Thus, if Q n is total consumption in n years, in billions of barrels, Q n = 27 r + 27 r 2 + 27 r 3 + · + 27 r n . Using the formula for the sum of a finite geometric series, we have Q n = 27 r 1 - r n 1 - r . Since the total reserves are 1148 billion barrels, in each case we find n making Q n = 1148 . (a) Using r = 0 . 995 , we solve 27(0 . 995) 1 - (0 . 995) n 1 - 0 . 995 = 1148 5373(1 - (0 . 995) n ) = 1148 1 - (0 . 995) n = 1148 5373 = 0 . 2137 (0 . 995) n = 0 . 7863 . Taking logs and using ln( A p ) = p ln A , n ln(0 . 995) = ln(0 . 7863) n = ln(0 . 7863) ln(0 . 995) = 48 . 0 years .
494 Chapter Eleven /SOLUTIONS (b) Using r = 1 . 025 , we have 27(1 . 025) 1 - (1 . 025) n 1 - 1 . 025 = 1148 1107((1 . 025) n - 1) = 1148 (1 . 025) n - 1 = 1148 1107 = 1 . 0370 (1 . 025) n = 2 . 0370 n ln(1 . 025) = ln(2 . 0370) n = ln(2 . 0370) ln(1 . 025) = 28 . 8 years. 3. After receiving the n th injection, the quantity in the body is 50 mg from the n th injection, 50(0 . 60) from the injection the previous day, 50(0 . 60) 2 from the injection two days before, and so on. The quantity remaining from the first injection (which has been in the body for n - 1 days) is 50(0 . 60) n - 1 . We have Quantity after n th injection = 50 + 50(0 . 60) + 50(0 . 60) 2 + · · · + 50(0 . 60) n - 1 . (a) The quantity of drug in the body after the 3 rd injection is Quantity after 3 rd injection = 50 + 50(0 . 60) + 50(0 . 60) 2 = 98 mg . Alternately, we could find the sum using the formula for a finite geometric series with a = 50 , r = 0 . 60 , and n = 3 : Quantity after 3 rd injection = 50(1 - (0 . 60) 3 ) 1 - 0 . 60 = 98 mg . (b) Similarly, we have Quantity after 7 th injection = 50 + 50(0 . 60) + 50(0 . 60) 2 + · · · + 50(0 . 60) 6 . We use the formula for the sum of a finite geometric series with a = 50 , r = 0 . 60 , and n = 7 : Quantity after 7 th injection = 50(1 - (0 . 60) 7 ) 1 - 0 . 60 = 121 . 5 mg . (c) The steady state level is the long term amount of drug in the body if injections are continued indefinitely. Right after an injection is given, we have Steady state level = 50 + 50(0 . 60) + 50(0 . 60) 2 + · · · .