Question

Use the formula for the \(n\) th partial sum of a geometric series $$\sum_{i=0}^{n-1} a r^{i}=\frac{a\left(1-r^{n}\right)}{1-r}$$ You go to work at a company that pays \(\$ 0.01\) for the first day, \(\$ 0.02\) for the second day, \(\$ 0.04\) for the third day, and so on. If the daily wage keeps doubling, what would your total income be for working (a) 29 days, (b) 30 days, and (c) 31 days?

Short answer

The total income for working 29 days is approximately 5368.71 USD, for 30 days is approximately 10737.42 USD, and for 31 days is about 21474.83 USD.

Step-by-step solution

Identify the parameters

Identify the first term a and the common ratio r in the geometric series. Here, a = 0.01 USD (the wage on the first day) and r = 2 (each day's wage is double the previous day's).

Apply the formula for a 29-day work period

Plug the parameters into the geometric series partial sum formula and calculate the sum for n = 29. This gives the total income for working 29 days: \[S = \frac{a(1 - r^n)}{1 - r} = \frac{0.01(1 - 2^{29})}{1 - 2}.\]

Apply the formula for a 30-day work period

Repeat this process for n = 30 to find the total income for working 30 days: \[S = \frac{a(1 - r^n)}{1 - r} = \frac{0.01(1 - 2^{30})}{1 - 2}.\]

Apply the formula for a 31-day work period

Finally, repeat the process once more for n = 31 to find the total income for working 31 days: \[S = \frac{a(1 - r^n)}{1 - r} = \frac{0.01(1 - 2^{31})}{1 - 2}.\]