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Interesting series question - the notation saves us!

For all real numbers, n, calculate:

$$\sqrt[3]{\dfrac{(1\cdot2\cdot4)+(2\cdot4\cdot8)+...+(n\cdot2n\cdot4n)}{(1\cdot3\cdot9)+(2\cdot6\cdot18)+...+(n\cdot3n\cdot9n)}}$$

for n=2022.

Not as bad as it looks, I promise! Check out the solution HERE.

Flex your trig identity muscles with this tricky trig limit!

An advanced factoring trick for a tough limit problem