PreCal

You have 1000 coins. Start by placing one coin in its own "stack." Next to the first stack, make a stack of two coins. Then make a stack of four coins and continue creating stacks of coins, each stack twice as high as the previous one.
Now I figured it would be 13 stacks by doing it out longhand.
What formula would you use to find this?

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  1. that is the sum of a geometric series
    1000 = sum from n = 1 to n = n of
    a + ar + a r^2 + a r^3
    where a = 1
    and r = 2
    the sum of the first n terms is
    a (1-r^n) /(1-r)
    so here
    1000 = 1 (1-2^n)/(1-2)
    -1000 = 1 - 2^n
    2^n = 1001
    n log 2 = log 1001
    n = 3/.301 = about 10

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  2. I don't think 10 is correct.
    I did it long hand
    1,2,4,8,16,32,64,128,256,512,1024,2048,
    4096 = 13 stacks.

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  3. Your terms are
    1, 2 , 4 , 8
    or
    1 ,2^1 , 2^2 , 2^4 ...... 2^12

    that sum is
    2^n = 10,001
    or
    n (.301) = 4
    n = 4/.301 = 13

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  4. In other words you either have a typo and you mean 10,000 coins, not 1,000 coins, or I am very confused

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  5. notice the following pattern

    2^0 = 1
    2^0 + 2^1 = 3 = 2^2 - 1
    2^0 + 2^1 + 2^2 = 7 = 2^3 - 1 , 3 terms in sum
    2^0 + 2^1 + 2^2 + 2^3 = 15 = 2^4 - 1 , 4 terms in sum
    ...
    2^0 + 2^1 + ... + 2^9 = 1023 = 2^10 - 1

    (the sum of n powers of 2 starting with 2^0 or 1, will be 2^(n+1) - 1

    so 11 terms give you a sum of 1023
    and 10 terms give you a sum of 511

    So as Damon calculated,
    9 stacks are not enough, using only 511 coins,
    and 10 stacks will leave the last stack short since you need 1023 to make 10 stacks

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