# How to derive a formula for the first n powers of an integer?

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.

• Probably you mean the sum of a geometric progression: en.wikipedia.org/wiki/Geometric_progression – Dennis Gulko Mar 14 '12 at 23:25
• 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. – user93200 Mar 14 '12 at 23:32

## 4 Answers

OK, here's a picture.

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 $$1 + r + \ldots + r^n = \frac{r^{n+1}-1}{r-1}$$

• Perhaps the widths of the rectangles should be $1/(r-1)$, so the sum of the areas of the glue rectangles is $1+r+\ldots r^n$. – Robert Israel Mar 15 '12 at 21:47
• 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. – user93200 Mar 15 '12 at 23:52

Do you mean you want a formula for $x + x^2 + \ldots x^n$? Look up "geometric series".

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 $$S(n) = 2^{n+1} - 2.$$

• I was looking rather for a way to easily see why the formula is true. Something like a picture proof or the proof that the sum of the numbers from 1 to n is n(n+1)/2 on maths.surrey.ac.uk/hosted-sites/R.Knott/runsums/triNbProof.html. It should also easily generalize to other integers besides 2. Thank you for a proof anyway. – user93200 Mar 15 '12 at 0:14
• Another homework problem – Kirthi Raman Mar 19 '12 at 15:12

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 $2^n$.

• Could you elaborate or draw a picture? – user93200 Mar 15 '12 at 0:15