Nested radicals

Algebra Level 5

Let $\mathcal{R}$ be the set of all the $2^{10}=1024$ real numbers of the form: $\pm\sqrt{2\pm\sqrt{2\pm \sqrt{ 2\pm \cdots \pm \sqrt{2}}}},$ where root signs appear exactly $10$ times. If $\sum_{r\in \mathcal{R}}\frac{1}{25+r}=\frac{a}{b},$ where $a$ and $b$ are coprime positive integers. What are the last three digits of $a+b$ ?

The answer is 927.

1 solution

Joshua Prettyman
Feb 20, 2014

when adding up all the elements of R, pair them up so that at first you have 512 pairs of the form 1/(25+x) + 1/(25-x), x is one of the nest square roots. At the next stage we have 256 pairs of the form 50/(623+x) + 50/(623-x), this time x has only 9 root signs. because you have the difference of two squares, adding them up is easy. Unfortunately it gets really big! say n is the numerator and d is part of the denominator. At one stage we have n/(d+x) + n(d-x). so the initial value of (n,d) is (1,25). then (50, 623). In fact, (a,b) -> (2ab, b^2 - 2). Finding the denominator part is easy: the last three digits go 25, 623, 127, 127, 127, and then they've settled down. Th numerator part alternates ending in 50, 300, 200, 800, 200, 800, ... 10 is an even number so we want the 800. 800+127 gives us the 927 that we want.

The only issue with this solution is that you don't really show that a and b are coprime. Of course I could have missed something...

Abhay Parekh - 7 years, 3 months ago

Nested radicals

The answer is 927.

1 solution

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