If the daily wage keeps doubling, what will your total income be for working 31 days? (Help pls:)) )

Question

Suppose you go to a company that pays 0.05 for the first day, 0.1 for the second day, 0.2 for the third day and so on. If the daily wage keeps doubling, what will your total income be for working 31 days?

Total Income =_____

Answer 1

This is a series that can be constructed as follows:

Day Pay

1 0.05

2 0.10

3 0.20

4 0.40

. .

. .

. .

If the day#1 pay is used as the common factor, then:

Day Pay

1 0.05×(1)

2 0.05×(2)

3 0.05×(4)

4 0.05×(8)

. .

. .

n 0.05×2^(n-1)

The total pay will be the sum from day 1 to day 31:

Total = 0.05(1)+0.05(2)+0.05(4)+0.05(8)+...0.05×2^(n-1) = 0.05(1+2+4+8+...2^(n-1)) = 0.05 ∑2^(n-1) from n=1 to 31; this is a geometric series of the inifinite kind when multiplying and dividing by 2:

Total = 0.025∑2ⁿ from n=1 to 31.

The sum for this series is:

∑2ⁿ = 2(1-2ⁿ)/(1-2) = -2(1-2³¹) = 4294967294

Total = 0.025 × 4294967294 = 107374182.35

Answer 2

This is called a geometric series problem. To solve it, we need to use a very strange-looking formula:

To find the sum of a finite geometric series, use this formula:  

S_n=a₁(1−rⁿ) / 1−r

where r ≠ 1, n = the number of terms, a₁ = the first term and r = the common ratio. The common ratio is the constant factor between consecutive terms in the geometric series.

Here, the common ratio is 2 because the daily wage doubles each day. The number of terms is 30 days, not 31, because the worker gets 5 cents on day 1 and then the sequence of doubling of pennies occurs. The first term is .01 because we are starting on the day when the person got the penny.

So:

S₃₀ = .01 (1 - 2³⁰) / 1 - 2

S₃₀ = .01 (1 - 1,073,741,824) / -1

S₃₀ = 10,737,418.23

Don't forget to add in that nickel from the first day.

$10,737,418.23 + .05 = $10,737,418.28