So we know that Krista put 1 cent into her bank on Sunday. And we know that she doubled the amount of money she put in each day. So on Monday she put in $1*2=2$ cents. And on Tuesday she put in $2*2=4$ cents. So we can say that the amount of money she put in each day was $2^n$ where n is the number of days since Sunday. So on the 7th day, she put in $2^6=64$ cents. And on the 14th day she put in $2^13=8192$ cents. Which is $81.92. So on the 13th day, she put in $2^12=4096$ cents. Which is $40.96 And the total amount of money she had must have been greater than 2 dollars on the 13th day. She has $2^0+2^1+2^2+...+2^n=2^{n+1}-1$. We can say that $M=1+2+4+...+2^n$ Which is the sum of a geometric series. Right. So we can use the formula for the sum of a geometric series to find the amount of money in her bank on the nth day. The sum is given by $S_n=\frac{1-2^{n+1}}{1-2}$ So the amount of money in her bank on the nth day is $M=\frac{1-2^{n+1}}{1-2}$ We need to find the smallest n such that $M>2$ So $\frac{1-2^{n+1}}{1-2}>2$ Ok, let's try that again. $\frac{1-2^{n+1}}{1-2} > 200$ $\frac{1-2^{n+1}}{-1} > 200$ $1-2^{n+1} < -200$ $2^{n+1} > 201$ So $n+1 \geq 8$ So $n \geq 7$ So the smallest value of n that satisfies the inequality is 7 So on the 7th day, which is $\text{Sunday}$, the total amount of money in her bank first exceeded $\$2$.


Krista put 1 cent into her new bank on a Sunday morning.  On Monday she put 2 cents into her bank.  On Tuesday she put 4 cents into her bank, and she continued to double the amount of money she put into her bank each day for two weeks.  On what day of the week did the total amount of money in her bank first exceed $\$2$?