TITLE: Power term calculated from preceding terms
QUESTION [9 upvotes]: 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?

REPLY [2 votes]: 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.