Stuck on this please help!!

Question

Nathan has 80 stamps in his collection. He adds 1 stamp to it today. Each day he plans to add twice the number of stamps as the previous day. If he keeps adding stamps at this rate for n days, which recursive function represents the number of stamps he has on any day in the future?

Answer 1

Again, no disrespect to Andrew's answer as it is correct.  However I doubt it is the apprpriate solution for a student in an algebra I class.

  

if you notice 

 

2^0 (two to the zero power)+ 2^1 (to the first power) + 2^2  ...and so on has an interesting pattern.

  1                                      +  2                                +  4     = 7  which is exactly 1 less than 2^3 or 8

  

similarly 7=8 = 15 or exactly 1 less than 2^4 or 16. thgis is always the case so the following formula give the same result as the other solution but avoids using pre-calc/calc methodology.

 

if n is the number of days, then day 1 is day 1 

so the number added on the nth day =                   2^(n-1)             day 1   add   2^(1-1)      = 2^0     = 1     

                                                                                                   day 2   add   2^(2-1)      = 2^1     = 2

since the sum of the prior days added pennies is always 1 less than the number added on the given nth day, 

the total added pennies can be expressed then by   2 * (2^(n-1)) - 1   and since   2 * (2^(n-1)) is the same as 2^n

                                                                           2 * (2^(n-1)) - 1   is simply       2^n - 1   added to the initial 80

 

 

ANSWER:  the total number of pennies on the nth day if starting with 80 and adding 1 on the first day, 2 on the second, 4 on the third, etc is  equal to

 

                                                     80 + (2^n  - 1)    where n is number of any given day starting with day 1

 

i.e  day 5 would have 80 + 1 + 2 + 4 + 8 + 16...which is            111

 

by the formula above:    80 +(2^n - 1) =                   80 + (2^5  -1)         =    80 +(32 - 1)        =        111                                                          

 

Answer 2

We start with 80 and we add 1 on the first day, 2 on the 2nd, 4 on the 3rd, etc..

 

Looking at the numbers being added

 

1   = 2⁰

2   = 2¹

4   = 2²

8   = 2³

16 = 2⁴

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.

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We start with 80 and we add 2ⁿ to it where n represents the number of days into the iteration

with n starting at 0

 

The function that shows how many stamps we have on any given day is

 

                _D

S = 80 + ∑ 2ⁿ       Where D is the number of days we are into the iteration

              ⁿ⁼⁰

NOTE:  This starts at day 0 which would be S = 80 + 2⁰ = 80 + 1 = 81 stamps

 

Example:  To find how many stamps we have on D = 4  we would have

S =  80+2⁰+2¹+2²+2³+2⁴

   =    80+1+2+4+ 8+16

     = 111 stamps