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(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!

Featherball
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    Are you familiar with geometric progressions? – Brian M. Scott Oct 31 '16 at 18:04
  • If you add a link to the duplicate question, someone may be able to more easily explain it, rather than having to write up a full working, if we know what has already been said... – JAP Oct 31 '16 at 18:05
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    "I can't ask a teacher yet... I need to work this out myself!" so... you asked it here, where many of the users are teachers themselves or are aspiring teachers... – JMoravitz Oct 31 '16 at 18:07
  • The original statement said to ask a teacher for the number, the user is asking for an explanation of a more mathematical approach, rather than just the answer, though I agree likely a teacher would give that as well, if asked – JAP Oct 31 '16 at 18:10
  • This question seems to give multiple approaches for the specific case of $r=2$, and the things which it has been labeled as a duplicate of have even more general results such as this one. – JMoravitz Oct 31 '16 at 18:14
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    I do think the shortest/easiest explanation is the one where you think in terms of base2 and you realize the number you end up with is $\underbrace{1111\dots 11_2}{64~~\text{ones}}$ which is just one less than the number $\underbrace{1000\dots 00_2}{1~\text{one and}~64~\text{zeroes}}$. – JMoravitz Oct 31 '16 at 18:20
  • I was thinking something more along the lines of:

    $\sum_{n=1}^{64}2^n$

     – JAP Oct 31 '16 at 18:21
  • Ah, well... Yeah... I didn't want to look silly in front of my teacher in case the answer was obvious. @JMoravitz – Featherball Oct 31 '16 at 18:47
  • Why has my question got a downvote? – Featherball Oct 31 '16 at 18:47
  • perhaps either because it still has not been made clear what you disliked about earlier posts and why you thought they were not thorough enough, thereby not making it clear what exactly you want from us. Or perhaps it is due to a lack of research effort on your own part, as this problem is very well known and understood and a cursory search for the phrase "geometric sum" should give you all the information you need. – JMoravitz Oct 31 '16 at 18:48

2 Answers2

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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 $$S=2^{64}-1$$

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

PascExchange
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  • Really good answer, thanks. What do you mean by 'the old one difference'? – Featherball Oct 31 '16 at 18:35
  • I'm sorry for being somewhat unclear. By the "old" one I mean $S$ and by the "new" one $2S$. I hope my change in the initial answer makes it more clear. – PascExchange Oct 31 '16 at 18:37
  • Ah, thanks so much. I might wait a bit before I accept though. I want to see if anyone else makes any interesting answers. – Featherball Oct 31 '16 at 18:42
  • @DanielCann Thank you. I added a further comment, to make you realize that this also holds in more generality (maybee you'll want to make your Chess-board bigger or even not only double the amount of rice-corns ;) – PascExchange Oct 31 '16 at 18:52
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observe $1+2=4-1=2^2-1$, $1+2+4=8-1=2^3-1$, $1+2+4+8=16-1=2^4-1$ can you prove this pattern? what does this pattern predict the answer to be?

shai horowitz
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