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Arithmetic sequences change additively by a constant difference; geometric sequences change multiplicatively by a constant ratio. sooo.....

 

Mr. Wiggins gives his daughter Celia two choices of payment for raking leaves:

Two dollars for each bag of leaves filled,

She will be paid for the number of bags of leaves she rakes as follows: two cents for filling one bag, four cents for filling two bags, eight cents for filling three bags, and so on, with the amount doubling for each additional bag filled.

 

If Celia rakes enough to five bags of leaves, should she opt for payment method 1 or 2? What if she fills ten bags of leaves?




How many bags of leaves would Celia have to fill before method 2 pays more than method 1?

DarkCalculis  Sep 25, 2017
edited by DarkCalculis  Sep 25, 2017
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 #1
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5 bags

 

Option 1....she gets  5 * 2  = $10

 

Option 2....she gets  .02 +.04 +.08 + .16 + .32  = 62 cents

 

 

10 bags

 

Option 1  .....she gets  10 * 2   = $20

 

Option 2.... using  the sum of a geometric series....we have that

 

S  = a1 [ 1 - r^n] / [ 1 - r]

 

S  is the sum      a1  is the first term  = 2 (cents)    

r is the common ratio between terms = 2     and n is the number of bags

 

So ......she gets   2 [ 1 - 2^10]  / [ 1 - 2 ]  = 2046 cents   =$20.46

 

So......Option 1 is better for 5 bags, but Option 2 is better for 10 bags

 

 

How many bags of leaves would Celia have to fill before method 2 pays more than method 1?

 

Note that when she fills 9 bags under Option 1  she gets  9 * 2  = $18

And when she fills 9 bags under Option 2 she gets  1022 cents  = $10.22

 

So...it appears that Option 2 is better than  Option 1 after she fills 10 bags  [or more ]

 

 

cool cool cool

CPhill  Sep 25, 2017

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