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How to derive a formula for the first n powers of an integer? In particular, sum of 2^n? I'm looking for a proof that not only utilizes algebraic manipulation but is also easily seen visually.

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    Yes, I think this is a specific type of geometric progression. However, I'm looking for a proof that not only utilizes algebraic manipulation but is also easily seen visually.2012-03-14

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OK, here's a picture.

enter image description here

The green rectangles have heights $1, r, r^2, \ldots r^n$ where in this case $n=7$. Each blue rectangle on the left has $r-1$ times the height of the green rectangle below it, so the green and blue together have height $r$ times the green alone. The sum of the lengths of the blue rectangles on the left is thus $r-1$ times the sum of the lengths of the corresponding green rectangles, i.e. $(r-1)(1 + r + \ldots + r^n)$. But on the right we see that the sum of the lengths of the blue rectangles is $r^{n+1} - 1$. So $(r-1)(1 + r + \ldots + r^n) = r^{n+1}-1$, or $1 + r + \ldots + r^n = \frac{r^{n+1}-1}{r-1}$

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    I appreciate this. Could you read it over and insert nouns and otherwise edit it for clarity? E.g., is "height" between "has" and "r−1"? "Each blue rectangle on the left has r−1 times the height of the green rectangle below it, so the green and blue together have height r times the green alone." I couldn't understand why this is so until I wrote (r−1)g+g=gr-g+g=gr. How is that evident from the picture? If the length of r was marked perhaps it'd be easier to see.2012-03-15
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Do you mean you want a formula for $x + x^2 + \ldots x^n$? Look up "geometric series".

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We'll show that $ \sum_{i=1}^{n} 2^i= 2^{n+1} - 2.$ Denote this sum by $ S(n) = 2^1 + 2^2 + \ldots + 2^{n-1} + 2^n \tag{1}$ On the RHS, factor out a $2$, we get $ S(n) = 2(1 + 2^1 + \ldots + 2^{n-1})$ But from $(1),$ we have $ 2^1 + \ldots + 2^{n-1} = S(n) - 2^n.$ So $ S(n) = 2(1 + \underbrace{2^1 + \ldots + 2^{n-1}}_{S(n) - 2^n})\\ S(n) = 2(1+ S(n)-2^n)\\ S(n) = 2 + 2S(n) -2^{n+1} $ Finally $ S(n) = 2^{n+1} - 2.$

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    Another homework problem2012-03-19
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The geometric series for \frac{1}{2} is easily seen by dividing a $1\times1$ square. You can get the one u want by multiplying by $2^n$.

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    Could you elaborate or draw a picture?2012-03-15