Question
In a $\triangle\text{ABC},$ if $\cos\text{A}=\frac{\sin\text{B}}{2\sin\text{C}'}$ then show that c = a.
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$\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \ldots$
$\frac{1 !}{n !}=\frac{1 !}{4 !}-\frac{4}{5 !}$