Rearranging the terms of the series into the usual "descending order" for polynomials, we get a series expansion of:

ax^{n-1 +}........ax + a

A basic property of polynomials is that if you divide xⁿ – 1 by x – 1, you'll get:

xⁿ^–1 + xⁿ^–2 + ... + x³ + x² + x + 1

That is:

a(xⁿ^–1 + xⁿ^–2 + ... + x³ + x² + x + 1) = a(xⁿ-1)/(x-1)

The above derivation can be extended to give the formula for infinite series, but requires tools from calculus. For now, just note that, for | r | < 1, a basic property of exponential functions is that rⁿ must get closer and closer to zero as n gets larger. Very quickly, rⁿ is as close to nothing as makes no difference, and, "at infinity", is ignored. This is, roughly-speaking, why the rⁿ is missing in the infinite-sum formula.

find f'(x) of (x^2) + 3x + 2.

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