Using geometric series, why is the polynomial $1 + x + x^{2} + x^{3}$ $1+x+x^2+x^3$ be written as $\frac{x^{4} - 1}{x - 1}$ $(x^4-1)/(x-1)$ assuming x does not equal 1?

Question

3751 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-14T23:28:27

This is a series with first term $1$ $1$ , common ratio $x$ $x$ and $4$ $4$ terms.

The formula for sum of a geometric series is $s_{n} = \frac{a (1 - r^{n})}{1 - r}$ $s_n = (a(1 - r^n))/(1 - r)$ .

$s_{4} = \frac{1 (1 - x^{4})}{1 - x}$ $s_4= (1(1 - x^4))/(1 - x)$

$s_{4} = \frac{1 - x^{4}}{1 - x}$ $s_4 = (1 - x^4)/(1 - x)$

$s_{4} = \frac{x^{4} - 1}{x - 1}$ $s_4 = (x^4 - 1)/(x- 1)$

Proved!!

Hopefully this helps!

Using geometric series, why is the polynomial 1+x+x2+x31+x+x^2+x^3 be written as x4−1x−1(x^4-1)/(x-1) assuming x does not equal 1?