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Algebra Difficulty 5.2 AIME, harder Find the answer

Is the number (1+12)(1+14)(1+16)…(1+12018)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{4}\right)\left(1+\frac{1}{6}\right) \ldots\left(1+\frac{1}{2018}\right) greater than, less than, or equal to 50?

A number or a short expression. Spacing and $ signs are ignored.

Solution

Call the expression SS. Note that (1+12)(1+14)(1+16)…(1+12018)<(1+11)(1+13)(1+15)…(1+12017)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{4}\right)\left(1+\frac{1}{6}\right) \ldots\left(1+\frac{1}{2018}\right)<\left(1+\frac{1}{1}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{5}\right) \ldots\left(1+\frac{1}{2017}\right). Multiplying these two products together, we get (1+11)(1+12)(1+13)…(1+12018)=21⋅32⋅43⋯20192018=2019\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \ldots\left(1+\frac{1}{2018}\right) = \frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdots \frac{2019}{2018} = 2019. This shows that S2<2019⟹S<2019<50S^{2}<2019 \Longrightarrow S<\sqrt{2019}<50 as desired.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.