Difficult G.P. Problem from a Difficult Math Book

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Homework Statement

15. The series $$a+ar-ar^2-ar^3+ar^4+ar^5-ar^6-ar^7+...,$$
where $a>0$, has its $k$^th term, $T_k$, defined by
$$T_k=ar^{k-1}$$ if $k$ is of the form $4p-3$ or $4p-2$ and
$$T_k=-ar^{k-1}$$ if $k$ is of the form $4p-1$ or $4p,$
where p is a positive integer. By rewriting the series as the sum of two geometric series, or otherwise, prove that the sum to $4n$ terms of the series is $$\frac{a(1+r)(1-r^4n)}{(1+r^2)}.$$

Homework Equations

$$S_n=\frac{a(1-r^n)}{1-r}$$

The Attempt at a Solution

$$a+ar-ar^2-ar^3+ar^4+ar^5-ar^6-ar^7+...$$
$$=(a+ar+ar^4+ar^5+...)-(ar^2+ar^3+ar^6+ar^7+...)$$
$$=(a+ar^4+ar^8+...)+(ar+ar^5+ar^9+...)-(ar^2+ar^6+ar^10+...)-(ar^3+ar^7+ar^{11}+...)$$
$$={}^1 S_n +{}^2 S_n +{}^3 S_n +{}^4 S_n$$
$$=\frac{a(1-(r^{4n})}{1-r^4}+\frac{ar(1-(r^{4n})}{1-r^4}-\frac{ar^2(1-(r^{4n})}{1-r^4}-\frac{ar^3(1-(r^{4n})}{1-r^4}$$
$$=a(1+r+r^2+r^3)\frac{a(1-r^{4n})}{(1-r^4)}$$
and now I am stUCK in the mUCK YUCK! I don't know how to factorise $1+r+r^2+r^3$.

 

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Try grouping terms like this: (1 + r) + (r² + r³). 1 + r is then a common factor of both terms.