Author Archives: kevin

For integer $n>1$, find the value of $$\dfrac{\sum_{i=1}^{n^2-1}\sqrt{n+\sqrt{i}}}{\sum_{i=1}^{n^2-1}\sqrt{n-\sqrt{i}}}$$ Click here for the solution. Solution: $$\dfrac{\sum_{i=1}^{n^2-1}\sqrt{n+\sqrt{i}}}{\sum_{i=1}^{n^2-1}\sqrt{n-\sqrt{i}}}=1+\dfrac{\sum_{i=1}^{n^2-1}\sqrt{n+\sqrt{i}}-\sum_{i=1}^{n^2-1}\sqrt{n-\sqrt{i}}}{\sum_{i=1}^{n^2-1}\sqrt{n-\sqrt{i}}}$$ $=1+\dfrac{\sum_{i=1}^{n^2-1}(\sqrt{n+\sqrt{i}}-\sqrt{n-\sqrt{i}})}{\sum_{i=1}^{n^2-1}\sqrt{n-\sqrt{i}}}=1+\dfrac{\sum_{i=1}^{n^2-1}\sqrt{(\sqrt{n+\sqrt{i}}-\sqrt{n-\sqrt{i}})^2}}{\sum_{i=1}^{n^2-1}\sqrt{n-\sqrt{i}}}$ $=1+\dfrac{\sum_{i=1}^{n^2-1}\sqrt{2n-2\sqrt{n^2-i}}}{\sum_{i=1}^{n^2-1}\sqrt{n-\sqrt{i}}}=1+\dfrac{\sqrt{2}\sum_{i=1}^{n^2-1}\sqrt{n-\sqrt{n^2-i}}}{\sum_{i=1}^{n^2-1}\sqrt{n-\sqrt{i}}}$ $=1+\dfrac{\sqrt{2}\sum_{i=1}^{n^2-1}\sqrt{n-\sqrt{i}}}{\sum_{i=1}^{n^2-1}\sqrt{n-\sqrt{i}}}=\boxed{1+\sqrt{2}}$