Power term calculated from preceding terms

Question

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?

Answer 1

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

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,

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

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

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

Answer 2

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

Answer 3

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*}

Answer 4

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.