sequences and series - How do I solve the chessboard problem?

(Note that I am a high school student and bad at maths. Please explain your answers thoroughly.)

And the man asked for grains of rice. He wanted one grain of rice on the first square of the chess board, two grains on the second, four grains on the third, eight grains on the fourth, so on and so on. I suppose you could ask a mathematics teacher if you really wished to learn how many pieces the man had after that was done for every square of the chess board.

Okay, so in school we were having some moral assembly or whatever, I didn't listen, too busy thinking about a more mathematical way to solve the problem above than just doubling, adding, etc. I can't ask a teacher yet... I need to work this out myself!

I'm an idiot, so I wanted help, and I was hoping someone on mathematics SE could help me think of a good mathematical way of solving this. Suppose there are 64 squares on the chess board. Argh! Damn this, I just found out that its a duplicate. Oh well. The answers on the question have to little an explanation.

Okay, there's 64 squares on our imaginary chessboard, and I'm trying to work out the thing above in a more mathematical way than just one by one, doubling it over and over again. So, if there is a more mathematical way of doing this, could someone tell me what it is and explain it.

Thanks!

Answer

Let's write down in a mathematical fashion what we have. If you think a while about it, you'll realize that you want to calculate is the sum
$$S=\sum\limits_{n=0}^{63}2^n=1+2+4+\dots+2^{63}$$
We can quickly check that this is indeed true if we realize that for $n=0$ we get $2^0=1$, followed by $2^1=2$, etc. The fact that the sum only goes up to $63$ is because if you start counting at $0$ you already spelled $64$ numbers when reaching $63$. How can we now calulate this sum? It requires a small trick namely to realize that if we compare the sum given by
$$2\cdot S=2\cdot \sum\limits_{n=0}^{63}2^n=\sum\limits_{n=1}^{64}2^n$$
and $S$, their difference is just given by
$$2S-S=2^{64}-1$$
Using this and just rewriting $2S-S=S$ we find the final result
$$S=2^{64}-1$$

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Additional comment:

We can also generalize this approach, which goes under the name of a geometric series. Consider a series of numbers which starts with $1$ and is constructed by just multiplying with a constant factor $q$. We further do this procedure $N$ times. The corresponding sum is than given by
$$S=\sum\limits_{n=0}^{N-1}q^n$$
Using the same procedure as above we find
$$q\cdot S=\sum\limits_{n=1}^N q^n$$
and therefore
$$q\cdot S-S=q^N-1$$
This time we have to factor $S$ out of the left hand term and find $S(q-1)$. If we now devide the equality by $(q-1)$ we find the more general result

$$S=\frac{q^N-1}{q-1}$$