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Question 6
If $$\left(1 + \frac{1}{2}\right)\left(1 + \frac{1}{4}\right)\left(1 + \frac{1}{6}\right)\left(1 + \frac{1}{8}\right)\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{5}\right)\left(1 - \frac{1}{7}\right) = 1 + \frac{1}{x}$$, then what is the value of x?
Solution
Now we know $$1+\frac{1}{2}=\frac{3}{2}$$ and $$1-\frac{1}{3}=\frac{2}{3}$$
Therefore $$\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)=1$$
similarly multiplying all we get $$\frac{9}{8}=1+\frac{1}{8}$$
so x=8
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