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- Feb 14, 2012
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Simplify \(\displaystyle \frac{\sum\limits_{k=1}^{99}\sqrt{10+\sqrt{k}}}{ \sum\limits_{k=1}^{99}\sqrt{10-\sqrt{k}}}\)
Hi jacks,My Solution:: I have Generalise the result.
Here we have to calculate $\displaystyle \frac{\sum_{k=1}^{n^2-1}\sqrt{n+\sqrt{k}}}{\sum_{k=1}^{n^2-1}\sqrt{n-\sqrt{k}}} = $
Let $\displaystyle A_{n} = \sum_{k=1}^{n^2-1}\sqrt{n+\sqrt{k}}$ and $\displaystyle B_{n} = \sum_{k=1}^{n^2-1}\sqrt{n-\sqrt{k}}$ , where $n>1$
Now $\left(\sqrt{n+\sqrt{k}}-\sqrt{n-\sqrt{k}}\right)^2 = 2n-2\sqrt{n^2-k}$
So $\left(\sqrt{n+\sqrt{k}}-\sqrt{n-\sqrt{k}}\right) = \sqrt{2}\cdot \sqrt{n-\sqrt{n^2-k}}$
So $\displaystyle \sum_{k=1}^{n^2-1}\left(\sqrt{n+\sqrt{k}}-\sqrt{n-\sqrt{k}}\right) = \sum_{k=1}^{n^2-1}\sqrt{2}\cdot \sqrt{n-\sqrt{n^2-k}}$
So So $\displaystyle \sum_{k=1}^{n^2-1}\left(\sqrt{n+\sqrt{k}}-\sqrt{n-\sqrt{k}}\right) = \sum_{k=1}^{n^2-1}\sqrt{2}\cdot \sqrt{n-\sqrt{k}}$
So $A_{n}-B_{n} = B_{n}\sqrt{2}$
So $A_{n} = B_{2}\left(1+\sqrt{2}\right)$
So $\displaystyle \frac{A_{n}}{B_{n}} = 1+\sqrt{2}$