# prove a equality

• Sep 12th 2010, 07:24 AM
Shanks
prove a equality
$\displaystyle \frac{\sum^{99}_{i=1}\sqrt{10+\sqrt{i}}}{\sum^{99} _{i=1}\sqrt{10-\sqrt{i}}}=1+\sqrt{2}$
• Sep 13th 2010, 11:16 AM
Opalg
Quote:

Originally Posted by Shanks
$\displaystyle \frac{\sum^{99}_{i=1}\sqrt{10+\sqrt{i}}}{\sum^{99} _{i=1}\sqrt{10-\sqrt{i}}}=1+\sqrt{2}$

Here is an outline of a proof. The full argument is quite long.

Step 1. If $\displaystyle 0<\theta<\pi/2$ then $\displaystyle \sqrt{1+\cos\theta}+\sqrt{1+\sin\theta} = (1+\sqrt2)\bigl(\sqrt{1-\cos\theta}+\sqrt{1-\sin\theta}\bigr).$

Reason: Let $\displaystyle t = \tan(\theta/2)$. Using the relations $\displaystyle \cos\theta = \tfrac{1-t^2}{1+t^2}$ and $\displaystyle \sin\theta = \tfrac{2t}{1+t^2}$, you can check that $\displaystyle \dfrac{\sqrt{1+\cos\theta}+\sqrt{1+\sin\theta}} {\sqrt{1-\cos\theta}+\sqrt{1-\sin\theta}} = \dfrac{1+\sqrt2+t}{1+(\sqrt2-1)t}.$ This is equal to $\displaystyle 1+\sqrt2$ because $\displaystyle \sqrt2-1 = \tfrac1{1+\sqrt2}.$

Step 2. If m, n are positive integers then

$\displaystyle \sqrt{\sqrt{m+n}+\sqrt m} + \sqrt{\sqrt{m+n}+\sqrt n} = (1+\sqrt2)\bigl(\sqrt{\sqrt{m+n}-\sqrt m} + \sqrt{\sqrt{m+n}-\sqrt n}\bigr).$

Reason: In Step 1, let $\displaystyle \cos\theta = \sqrt{\tfrac m{m+n}}$. Then $\displaystyle \sin\theta = \sqrt{\tfrac n{m+n}}$.

Step 3. For $\displaystyle 1\leqslant m\leqslant49$ and $\displaystyle n = 100 -m$, $\displaystyle \sqrt{10+\sqrt m} + \sqrt{10+\sqrt n} = (1+\sqrt2)\bigl(\sqrt{10-\sqrt m} + \sqrt{10-\sqrt n}\bigr).$ Also, $\displaystyle \sqrt{10+\sqrt 50} = (1+\sqrt2)\sqrt{10-\sqrt 50}.$

Reason: Follows immediately from Step 2 because $\displaystyle 10 = \sqrt{100}$.

Step 4. $\displaystyle \displaystyle\sum_{n=1}^{99}\sqrt{10+\sqrt{n}} = (1+\sqrt2)\sum_{n=1}^{99}\sqrt{10-\sqrt{n}}$.

Reason: Just add up all the relations in Step 3.

Finally, divide by $\displaystyle \sum_{n=1}^{99}\sqrt{10-\sqrt{n}}$ to get the result.