Library / /191 of 241

, 2024

Algebra Difficulty 2.5 Junior Find the answer Canada

Suppose that $\$\sqrt{\dfrac{1}{2} \times \dfrac{2}{3} \times \dfrac{3}{4} \times \dfrac{4}{5} \times \cdots \times \dfrac{n-1}{n}} = \dfrac{1}{8}\$.$ (The expression
under the square root is the product of $n-1$ fractions.) The value of $n$ is

Pick one

Solution

Since $\sqrt{\dfrac{1}{2} \times \dfrac{2}{3} \times \dfrac{3}{4} \times \dfrac{4}{5} \times \cdots \times \dfrac{n-1}{n}} = \dfrac{1}{8}$ then squaring both sides,
we obtain $\dfrac{1}{2} \times \dfrac{2}{3} \times \dfrac{3}{4} \times \dfrac{4}{5} \times \cdots \times \dfrac{n-1}{n} = \dfrac{1}{64}$ Simplifying the left side, we
obtain $\dfrac{1 \times 2 \times 3 \times 4 \times \cdots \times (n-1)}{2 \times 3 \times 4 \times 5 \times \cdots \times n} = \dfrac{1}{64}$ or $\dfrac{1 \times \left(2 \times 3 \times 4 \times \cdots \times (n-1) \right)}{\left(2 \times 3 \times 4 \times \cdots \times (n-1) \right) \times n} = \dfrac{1}{64}$ and so $\dfrac{1}{n} = \dfrac{1}{64}$ which means
that $n = 64$ .

Want a route through all this instead of an archive? The track puts 2,444 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.