is (xn - 1)/(x - 1). One of my beliefs (as also shown by that post) is that you should never take a formula at its face value. Here I quickly attempt to show why this particular formula should be correct using a very commonly used proof. Let us assume that S denotes the desired sum. That is:
If we multiply both sides of this equation by x, we get:
If we subtract each side of the first equation from the corresponding side of the second equation we get:
This simplifies to:
as we had claimed.
By the way, while attending a conference call last night where it was particularly hard to concentrate on what was being discussed, I began to doodle and stumbled upon a simple result which might have been interesting about a millennium ago and is probably just an idle curiosity now. I am also sure it is contained somewhere in our high school text books for mathematics, but I cannot recall seeing it before. (Despite what might be indicated by these blog posts, I do not have much of an interest in mathematics itself and have never pursued it for its own sake.)
Anyways, looking at the factorisation of (x2 - y2) into (x - y)×(x + y) and (x3 - y3) into (x - y)×(x2 + x×y + y2), I wondered if (x - y) is always a factor of (xn - yn). By playing around a little bit with the terms, I could write:
confirming my hypothesis. If you set y=1, this reduces to:
which can be rewritten as:
which is the same formula that we derived just a while back for calculating the sum of a geometric series!