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The question seems trivial at first, but I have problems wrapping my head around this.

Consider a business that starts on day 0 with 1000 customers each paying 6 dollar for 6 months in advance. There are no partial payments or anything. The payment interval stays the same fixed to 6 months. Now, every day, the business gains 3 more customers, each paying 6 dollar for 6 months. After 6 months, the initial 1000 customers pay again their 6 dollar for the following interval. The day after, there are 3 recurring payments plus 3 new payments and so on.

The question is: How much money does the business have on hand on a given day x?

Let's assume for simplicity, that 6 months = 180 days. Some calculations:

 Day 0: 1000 * 6$ Day 1: (1000 + 3*1) * 6$ Day 100: (1000 + 3*100) * 6$ Day 180: (1000*2 + 3*180) * 6$ Day 181: (1000*2 + 3*181 + 3*1) * 6$ 

Can someone help me figuring out the formula to calculate the money available on a given day X?

2 Answers 2

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Using conventions from a previous answer:

$ 6\times\left(10^3\left(\left\lfloor \frac{N}{180}\right\rfloor + 1\right) + 3 \left( N \left(\left\lfloor \frac{N}{180}\right\rfloor + 1\right) - \sum_{n=0}^{\lfloor \frac{N}{180}\rfloor} 180n \right)\right)$

For the Nth day, where $\lfloor\cdot \rfloor$ is the floor function.

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    Thanks. That's the solution I came up with as well, but I wasn't allowed to post because of newbie rules and didn't check again :)2011-04-13
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On day $d$, the business has

$6\times \left( 1000 \times (\left\lfloor \frac{d}{180}\right\rfloor + 1) + 3 \times (d \mod{180}) + \sum_{n=0}^{\lfloor \frac{d}{180}\rfloor} 3 \times 180 \times n \right)$