Question 726849
This is a geometric sequence with a factor of 2: 2, 4, 8, 16...<P>
The sum of a geometric series is {{{a(1)* (1-r^n)/(1-r)}}} where a(1) is the first term, r is the common factor, and n is the nth term (the number of the term that = the sum).  Set this =  500 and solve for n.  The n needed (the numbered term) will be greater than that n.<P>
a(1) = 2  and r=2.  The desired sum is 500.
{{{500 = 2*(1-2^n)/(1-2) }}}<P>
{{{-500 = 2(1-2^n)}}}<P>
{{{-250 = 1-2^n}}}<P>
{{{2^n = 251}}}<P>
Recall that {{{x^n=y}}} when log (base x)y = n so log(base 2) 251 = n = approx. 7.97<P>
Therefore the sum of the first 7.97 terms is 500, so the sum exceeds 500 with the 8th term or 8th day.<P>
You could have also solved this by writing out the amount he saved each day, and adding:  2, 4, 8, 16, 32, 64, 128, 256 (the sum is $510, which exceeds $500 on the 8th day.)
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