Question
Prove that: $\Big(2^{\frac14}.4^{\frac18}.8^{\frac{1}{16}}.16^{\frac{1}{32}}\dots\infty\Big)=2$
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Find the middle term in the expansion of:
$\Big(\frac{2}{3}\text{x}-\frac{3}{2\text{x}}\Big)^{20}$
$\frac{\sec\text{x}-1}{\sec\text{x}+1}$