I discovered a new summation formula regarding the consecutive powers of a given base. I noticed, to start, that any given power of 2 could be written as a sum involving the previous powers of two- for example, 4-1 = 2+1, or 32-1 = 16+8+4+2+1, generalized in the formula Sum(i=0, n-1) {2ⁱ } = 2ⁿ - 1. My idea was to expand this formula to include the powers of any base.

So, I first asked- is it even possible to write every power of 3 as a sum of the terms preceding it? Instinct said yes, and with a little guesswork I noticed a pattern: 3-1 = 2(1), 9-1 = 2(1+3), 27-1 = 2(1+3+9), and so on. This led to the general formula, Sum(i=0, n-1) {3ⁱ} = (3ⁿ - 1)/2.]

Before I state the general formula for xⁿ, which you might be able to guess, I would like to note that there is a somewhat non-traditional proof of this kind of formula that I used to verify my results. I'll give an example using 3.

Our goal is to prove that 1+2(1+3+9+...+3^n-1) = 3ⁿ

Multiplying out and adding the 1 to the first term of the sum,

3+2(3)+2(9)+...+2(3^n-1) = 3ⁿ

Adding the first two terms of the LHS,

9+2(9)+2(27)+...+2(3^n-1) = 3ⁿ

Adding the first two terms of the LHS,

27+2(27)+...+2(3^n-1) = 3ⁿ

And so on. This domino process by which each preceeding sum adds with the next term to create a power of three can be quantified by this relationship: 3^k-1 + 2(3^k-1) = 3^k. This series of summations eventually ends with 3ⁿ.

A similar proof can be can be used to show that the formula for 2^n works, or for any consequent base. Here's the general formula: 1+(x-1)*Sum(i=0, n-1) {xⁱ} = xⁿ. With this formula, we can now see what the main hinge of the given proof really is- it is that the first "domino" in the series comes from 1+(x-1)(1) = x, which is exactly what is needed.

After creating this formula and remembering one of my older finds on the same topic, I had the desire to somehow combine the two finds and generate some cool new relationship. (The other formula came up with the same kind of result, a formula for xⁿ, though in that case the formula was a sum of polynomial values instead of exponential values.) I'll have to go back and see if there's any way to reconcile the two.