Question 1188981
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Day 1 = 3 nickels
Day 2 = 9 nickels
Day 3 = 27 nickels
Day 4 = 81 nickels
Each new day, we triple the previous day's amount. Example: 9*3 = 27 when jumping from day 2 to day 3.


Let n be a positive whole number (1,2,3,...)
For day n, we have 3^n nickels. 
Examples: 
Day n = 2 has 3^n = 3^2 = 9 nickels
Day n = 4 has 3^n = 3^4 = 81 nickels


As you probably can notice, the sequence 3,9,27,81,... is a geometric sequence. The first term is a = 3 and the common ratio is r = 3.


Use this formula to sum the first n terms of a geometric sequence
{{{S[n] = a*(1-r^n)/(1-r)}}}


For example, let's sum the first 4 days worth of nickels.
{{{S[n] = a*(1-r^n)/(1-r)}}}


{{{S[4] = 3*(1-3^4)/(1-3)}}}


{{{S[4] = 120}}}
Then confirm it by adding the first four items mentioned at the top of this page: 3+9+27+81 = 120
This example helps show the {{{S[n]}}} formula works. I encourage you to try other values of n.


The goal is to somehow get {{{S[n]}}} to be more than 700.
We can try n = 5
{{{S[n] = a*(1-r^n)/(1-r)}}}


{{{S[5] = 3*(1-3^5)/(1-3)}}}


{{{S[5] = 363}}}
which unfortunately doesn't work. If you tried n = 6, then you should find that {{{S[6] = 1092}}} which fits the goal. I'll leave those steps for you.


Sure enough, 
Day 1 = 3 nickels
Day 2 = 9 nickels
Day 3 = 27 nickels
Day 4 = 81 nickels
Day 5 = 243 nickels
Day 6 = 729 nickels
3+9+27+81+243+729 = 1092
which confirms {{{S[6] = 1092}}} is correct


<font color=red>Answer: 6 days</font>
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