Simplifying product of differences

sequences-and-series

Why didn't you even try for small numbers? If you had, you could have canceled out all the terms. ;)

The expression turns out to be

$\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5} \dots\dfrac{n-2}{n-1}.\dfrac{n-1}{n}$

Cancel out the terms and you're left with just $\frac{2}{n}$.

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