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Prove that $a + ar + ar^{2} + ...$(to n terms) = $\frac{a(1 - r^{n})}{1 - r}$ for $r \neq 1$
in Mathematics by Diamond (67,406 points) | 65 views

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The sum of the first n terms of the geometric sequence, in expanded form, is as follows:

a + ar + ar^2 + ar^3 + ... + ar^n–2 + ar^n–1

Polynomials written with their terms in descending order. Reversing the order of the summation above to put its terms in descending order, we get a series expansion of:

ar^n–1 + ar^n–2 + ... + ar^3 + ar^2 + ar + a

We can take the common factor of "a" out front:

a(r^n–1 + r^n–2 + ... + r^3 + r^2 + r + 1)

A basic property of polynomials is that if you divide x^(n – 1) by x – 1, you'll get:

x^n–1 + x^n–2 + ... + x^3 + x^2 + x + 1

This gives:

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

If we reverse both subtractions in the fraction above, we will obtain the following equivalent equation:

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

Applying the above to the geometric summation (by using instead of x), we get:

ar^{n-1} + ar^{n-2} + ... + ar^2 + ar + a

a(r^{n-1} + r^{n-2} + ... + r^2 + r + 1)

= a{1 - r^n}/{1 - r}  , proven

by Diamond (42,982 points)

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