Geometric Series for Catching Fish Each Day

Date: 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/