In my previous post, I said that the sum of the geometric series:

^{2}+ ... + x

^{n-1}

is (x

^{n}- 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:

^{2}+ ... + x

^{n-1}

If we multiply both sides of this equation by x, we get:

^{2}+ x

^{3}+ ... + x

^{n}

If we subtract each side of the first equation from the corresponding side of the second equation we get:

^{2}+ x

^{3}+ ... + x

^{n}- 1 - x - x

^{2}- ... - x

^{n-1}

This simplifies to:

^{n}- 1

which gives:

^{n}- 1)/(x - 1)

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 (x^{2} - y^{2}) into (x - y)×(x + y) and (x^{3} - y^{3}) into (x - y)×(x^{2} + x×y + y^{2}), I wondered if (x - y) is always a factor of (x^{n} - y^{n}). By playing around a little bit with the terms, I could write:

^{n}- y

^{n}= (x - y)×(x

^{n-1}+ x

^{n-2}×y + ... + x×y

^{n-2}+ y

^{n-1})

confirming my hypothesis. If you set y=1, this reduces to:

^{n}- 1

^{}= (x - 1)×(x

^{n-1}+ x

^{n-2}+ ... + x + 1

^{})

which can be rewritten as:

^{n}- 1)/(x - 1) = x

^{n-1}+ x

^{n-2}+ ... + x + 1

which is the same formula that we derived just a while back for calculating the sum of a geometric series!