Generating Functions and Polynomial Expansions Give a formula similar to:
$\frac{1-x^{m+1}}{1-x} = 1 + x + x^2 + ... + x^m$
For the following
(a) $1 + x^4 + x^8 + ... + x^{24}$
(b) $x^{20} + x^{40} + ... + x^{180}$
Workings
a. $1 + x^4 + x^8 + ... + x^{24}$
$g(x) = \frac{1-x^{25}}{1-x^4}$
b. $x^{20} + x^{40} + ... + x^{180}$
$g(x) = x^{20}(1 + x^{20} + ... + x^{160})$
$g(x) = x^{20}\frac{1-x^{161}}{1-x^{20}}$
I'm wondering if I am correct. Any help wil be appreciated.

You almost got it! But are just a little bit off. One way to check each of these is to introduce $y = x^{4}$ for a) and $z = x^{20}$ for b). This means $$1 + x^4 + x^8 + ... + x^{24} = 1+y+y^2+\ldots +y^6 \\ = \frac{1-y^{7}}{1-y}  \\ = \frac{1-x^{28}}{1-x^4}$$ and $$x^{20} + x^{40} + ... + x^{180} = z+z^2+\ldots +z^8 + z^9 \\ =(1+z+z^2+\ldots +z^8+ z^9)-1 \\ = \frac{1-z^{10}}{1-z}-1 \\ = \frac{1-z^{10}}{1-z}-\frac{1-z}{1-z} \\ = \frac{z(1-z^9)}{1-z} \\ = \frac{x^{20}(1-x^{180})}{1-x^{20}}$$
