# Math Help - Quickie #15

1. ## Quickie #15

Simplify: . $\left(1 + 2^{-\frac{1}{32}}\right)\left(1 + 2^{-\frac{1}{16}}\right)\left(1 + 2^{-\frac{1}{8}}\right)\left(1 + 2^{-\frac{1}{4}}\right)\left(1 + 2^{-\frac{1}{2}}\right)$

Edit: 34 views and no soltuion yet?

2. Finally found one that stumped all of you?

Let: . $X \;=\;\left(1 + 2^{-\frac{1}{32}}\right)\left(1 + 2^{-\frac{1}{16}}\right)\left(1 + 2^{-\frac{1}{8}}\right)\left(1 + 2^{-\frac{1}{4}}\right)\left(1 + 2^{-\frac{1}{2}}\right)$

Multiply both sides by $\left(1 - 2^{-\frac{1}{32}}\right)$

$\left(1 - 2^{-\frac{1}{32}}\right)X \;=\;\underbrace{\left(1 - 2^{-\frac{1}{32}}\right)\left(1 + 2^{-\frac{1}{32}}\right)}\left(1 + 2^{-\frac{1}{16}}\right)$ $\left(1 + 2^{-\frac{1}{8}}\right)\left(1 + 2^{-\frac{1}{4}}\right)\left(1 + 2^{-\frac{1}{2}}\right)$

. . . . . . . . . . . . . . $= \;\underbrace{\left(1 - 2^{-\frac{1}{16}}\right) \left(1 + 2^{-\frac{1}{16}}\right)}\left(1 + 2^{-\frac{1}{8}}\right)\left(1 + 2^{-\frac{1}{4}}\right)\left(1 + 2^{-\frac{1}{2}}\right)$

. . . . . . . . . . . . . . . . . . $= \;\underbrace{\left(1 - 2^{-\frac{1}{8}}\right)\left(1 + 2^{-\frac{1}{8}}\right)}\left(1 + 2^{-\frac{1}{4}}\right)\left(1 + 2^{-\frac{1}{2}}\right)$

. . . . . . . . . . . . . . . . . . . . . . $=\;\underbrace{\left(1 - 2^{-\frac{1}{4}}\right)\left(1 + 2^{-\frac{1}{4}}\right)}\left(1 + 2^{-\frac{1}{2}}\right)$

. . . . . . . . . . . . . . . . . . . . . . . . . . $= \;\underbrace{\left(1 - 2^{-\frac{1}{2}}\right)\left(1 + 2^{-\frac{1}{2}}\right)}$

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . $= \;1 - 2^{-1}$

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . $=\;\frac{1}{2}$

Therefore: . $X \;= \;\frac{\frac{1}{2}}{1 - 2^{-\frac{1}{32}}} \;=\;\frac{1}{2 - 2^{\frac{31}{32}}}

$

3. Thankyou! I was waiting forever for someone to do this!

4. Originally Posted by Soroban
Finally found one that stumped all of you?
Of course not! My calculator came up with 23.334023182557 a long time ago!

-Dan