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  1. #1
    Senior Member Shanks's Avatar
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    prove a equality

    \frac{\sum^{99}_{i=1}\sqrt{10+\sqrt{i}}}{\sum^{99}  _{i=1}\sqrt{10-\sqrt{i}}}=1+\sqrt{2}
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  2. #2
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    Quote Originally Posted by Shanks View Post
    \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 0<\theta<\pi/2 then \sqrt{1+\cos\theta}+\sqrt{1+\sin\theta} = (1+\sqrt2)\bigl(\sqrt{1-\cos\theta}+\sqrt{1-\sin\theta}\bigr).

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

    Step 2. If m, n are positive integers then

    \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 \cos\theta = \sqrt{\tfrac m{m+n}}. Then \sin\theta = \sqrt{\tfrac n{m+n}}.

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

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

    Step 4. \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 \sum_{n=1}^{99}\sqrt{10-\sqrt{n}} to get the result.
    Last edited by Opalg; September 13th 2010 at 01:40 PM. Reason: small correction
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