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, 2015

Algebra Difficulty 5.0 AIME Prove it Japan

Reduce the following expression into the form $a + b\sqrt{2}$ , where both $a$ and $b$ are rational numbers:
$\frac{(1 \times 4 + \sqrt{2})(2 \times 5 + \sqrt{2})\cdots(10 \times 13 + \sqrt{2})}{(2 \times 2 - 2)(3 \times 3 - 2)\cdots(11 \times 11 - 2)}$

Solution

$\boxed{11+5\sqrt{2}}$
$\text{For } k = 1, 2, \dots, 10, \text{ we have}$
$\frac{k(k+3)+\sqrt{2}}{(k+1)^2-2} = \frac{(k+1+\sqrt{2})(k+2-\sqrt{2})}{(k+1+\sqrt{2})(k+1-\sqrt{2})} = \frac{k+2-\sqrt{2}}{k+1-\sqrt{2}}$
$\text{Therefore, we obtain}$
$\begin{aligned} \frac{(1 \times 4 + \sqrt{2})(2 \times 5 + \sqrt{2}) \cdots (10 \times 13 + \sqrt{2})}{(2 \times 2 - 2)(3 \times 3 - 2) \cdots (11 \times 11 - 2)} &= \frac{3 - \sqrt{2}}{2 - \sqrt{2}} \cdot \frac{4 - \sqrt{2}}{3 - \sqrt{2}} \cdots \frac{12 - \sqrt{2}}{11 - \sqrt{2}} \\ &= \frac{12 - \sqrt{2}}{2 - \sqrt{2}} = 11 + 5\sqrt{2} \end{aligned}$

$\text{for}$ the desired answer.}

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