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My 11 year old son was playing around with powers of 3 (he's like that) and came up with an interesting pattern. We worked together to extend it and came up with this observation:

$$a^b = 1 + (a-1) \sum_{n=0}^{b-1} a^n$$

where 'a' and 'b' are integers of 1 or greater. I tried to prove it but I haven't done integrals for over 20 years and my results were way off, or maybe it just isn't true.

That said it looks so interesting that if it is true, I'd be suprised if it doesn't have a name.

Anyway, can anyone prove the RHS does in fact equal the LHS?

Swanny
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    This is just the well-known partial sum of a geometric series: see the fifth equation at http://mathworld.wolfram.com/GeometricSeries.html. – symplectomorphic Mar 01 '17 at 02:32
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    Good work you two! This is the same as $\sum_{n=0}^{b-1} a^n = \cfrac{a^b -1}{a-1}$ a.k.a. the sum of a geometric progression. $b$ must indeed be a positive integer, while $a$ can be any real number (including $1$ the way you have it written). – dxiv Mar 01 '17 at 02:34
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    Yes, just subtract $1$ from both sides and divide by $a-1$ to get the formula @dxiv posted above. (Your son is very young to be this interested in math! What was the highest power of $3$ he got to?) – Michael Wang Mar 01 '17 at 02:35
  • 10 years from now, let us know what new theorems your son has come up with – mrnovice Mar 01 '17 at 02:35
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    Thanks everyone. Some great answers here. Hard to pick the best one. I got fixated with integrals, but once I put that aside the answer seems obvious. I should have just attacked it like a series. Hopefully this will encourage him. I should dust off my old text books... uhm... or maybe find a good site. – Swanny Mar 01 '17 at 03:05
  • With regards to how high he got, he was only at 243 when he came to me with the idea (2 times sum of preceding terms + 1) and we used that to calculate the next two terms and verified them. Of course we already had the accumulated sum to work from. Then we went off to try 2 to the power and 4 to the power and then just got carried away. – Swanny Mar 01 '17 at 03:10
  • MathsJax is so cool by the way. Compliments to whoever came up with it. – Swanny Mar 01 '17 at 03:12

4 Answers4

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$$a^b = 1 + (a-1) \sum_{n=0}^{b-1} a^n$$

With a little rearrangement we get,

$$\frac{a^b-1}{a-1}= \sum_{n=0}^{b-1} a^n$$

So your question is really to prove the formula for a geometric sum. Here's one way to prove it,

$$a^n=\frac{a-1}{a-1}a^n=\frac{a^{n+1}}{a-1}-\frac{a^n}{a-1}$$

So that really we have,

$$\sum_{n=0}^{b-1} a^n=\sum_{n=0}^{b-1} \left(\frac{a^{n+1}}{a-1}-\frac{a^n}{a-1}\right)$$

Expand the right hand side, you'll see a lot cancels, we have a telescoping series. Anyways we have,

$$=\frac{1}{a-1} \sum_{n=0}^{b-1} (a^{n+1}-a^n)$$

$$=\frac{1}{a-1}(a^b-a^0)$$

$$=\frac{a^b-1}{a-1}$$

Ahmed S. Attaalla
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The right-hand side is a telescoping sum.

We obtain \begin{align*} 1+(a-1)\sum_{n=0}^{b-1}a^n&=1+\sum_{n=0}^{b-1}a^{n+1}-\sum_{n=0}^{b-1}a^n\\ &=1+\sum_{n=1}^ba^n-\sum_{n=0}^{b-1}a^n\\ &=1+a^b-1\\ &=a^b \end{align*}

Markus Scheuer
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Using the formula for the partial sums of a geometric series $$1+a+a^2+...+a^n=\frac{1-a^{n+1}}{1-a},$$ $$a^b=1+(a-1)\sum_{n=0}^{b-1}a^n=1+(a-1)\left(1+a+a^2+...+a^{b-1}\right)=1+(a-1)\frac{1-a^{b}}{1-a}$$ $$=1-(1-a^b)=a^b$$

garserdt216
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One nice way you can interpret this relationship is as a statement about numbers written in base $a$: If you have a number represented by a string of digits, each being the largest possible base-$a$ digit, and add 1, you get a power of $a$. For example: In base ten, if you have a string of 9's, adding 1 to that string 'propagates the carries' all the way past the high end, yielding an exact power of ten. So what your son has discovered is that that 'trick' works in any base.

PMar
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