### Warm-up:

Sequences

(care of mmerevise.co.uk)

### The Problem:

A billionaire donates to charity 1p. The next month they donate twice as much. Then the following twice as much again continuing to double the donation each month. After how many years will they run out of money?

-/|\-

### Solution:

We can start this solution to this problem with each months donation and add them up ...

1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 + 2048 + ...

So, after one year we've reach a total more than £4 which means we still have most of the billion pounds still remaining. This method is going to take a while.

Looking at the problem from the other end. The donator is going to give up at most half of their wealth in the last month. So this is looking more promising

50 000 000 000 + 25 000 000 000 + 12 500 000 000 + 6 250 000 000 + 3 125 000 000 + 1 562 500 000 + 781 250 000 + 390 625 000 + 195 312 000 + 97 656 250 + ...

Hence, in the last ten months the donator has given up most of their money. We're looking at a time to use up all the money as more than two years, but how much more?

We need a bit of A level maths to solve this one. It goes a bit like this ...

The sum of $n$ terms of a geometric sequences is given by

$${S}_{n}=\frac{a(1-{r}^{n})}{1-r}$$where ${S}_{n}$ is the total amount of wealth, $a$ is the initial donation, $r$ is the common ratio, and $n$ is the number of months donations were made.

Rearranging for $n$ ...

$${S}_{n}\left(1-r\right)=a(1-{r}^{n})$$ $$\frac{{S}_{n}\left(1-r\right)}{a}=1-{r}^{n}$$ $$1-\frac{{S}_{n}\left(1-r\right)}{a}={r}^{n}$$ $$ln(1-\frac{{S}_{n}\left(1-r\right)}{a})=ln{r}^{n}$$ $$ln(1-\frac{{S}_{n}\left(1-r\right)}{a})=nlnr$$ $$n=\frac{ln(1-\frac{{S}_{n}\left(1-r\right)}{a})}{lnr}$$The values of these are ${S}_{n}={10}^{11}$, $a=1$, $r=2$, and $n$ is to be found.

Substituting these values in we get ...

$$n=36.5412$$The billionaire would run out on money in just over three years with the last payment not being met in full.

### Further Reading:

This problem is a variation on the "Wheat and chessboard problem"

(care of en.wikipedia.org)