ben saved 1 penny on january 1, 2 pennies on january 2, 4 pennies on january 3, and 8 pennies on january 4. if ben continues to increase his saving at this rate, when will he have saved a total of $10 or more? show your work
.01
.02
.04
.08
.16
.32
.64
1.28
2.56
5.12
10.24
if you have 5,368,709.12 can anyone help with what is the interest rate and number of periods for this? thanks
To find out when Ben will have saved a total of $10 or more, we need to first calculate the total amount Ben saves each day starting from January 1.
On January 1, Ben saves 1 penny.
On January 2, Ben saves an additional 2 pennies, so now he has a total of 1 + 2 = 3 pennies.
On January 3, Ben saves an additional 4 pennies, so his total now becomes 3 + 4 = 7 pennies.
On January 4, Ben saves an additional 8 pennies, making his total savings 7 + 8 = 15 pennies.
Now we can see a pattern emerging: Ben saves 1 + 2 + 4 + 8 + ... pennies each day. This is a geometric sequence where the common ratio is 2 (each term is twice the previous one) and the first term is 1.
We can use the formula for the sum of a finite geometric series to find the total savings.
The formula is given by: Sn = a * (1 - r^n) / (1 - r), where Sn is the sum of the series, a is the first term, r is the common ratio, and n is the number of terms.
In this case, we need to solve the equation: 1 * (1 - 2^n) / (1 - 2) >= 1000 (since we want Ben to save $10 or more, which is equivalent to 1000 pennies).
Simplifying the equation, we have: (1 - 2^n) / (-1) >= 1000
Multiplying both sides by (-1), we get: 2^n - 1 >= -1000
Adding 1 to both sides, we have: 2^n >= -999
To solve this inequality, we take the logarithm (base 2) of both sides to isolate n:
log2(2^n) >= log2(-999)
n >= log2(-999)
Using a calculator, we find log2(-999) ≈ -9.9658 (rounded to 4 decimal places).
Since n represents the number of days that have passed, it cannot be negative. Therefore, we take the ceiling value of -9.9658, which is -9.
So, we conclude that Ben will have saved a total of $10 or more, after at least 10 days.