Question

If $S = \frac{\sum\limits_{n=1}^{99} \sqrt{10} + \sqrt{n}}{\sum\limits_{n=1}^{99} \sqrt{10} - \sqrt{n}}$ then the value of $\left( \frac{1}{S} - S \right)$ is

Answer 1

3/2

Answer 2

Let $N = \sum\limits_{n=1}^{99} (\sqrt{10} + \sqrt{n})$ and $D = \sum\limits_{n=1}^{99} (\sqrt{10} - \sqrt{n})$. $N = 99\sqrt{10} + \sum\limits_{n=1}^{99} \sqrt{n}$. $D = 99\sqrt{10} - \sum\limits_{n=1}^{99} \sqrt{n}$. Let $A = \sum\limits_{n=1}^{99} \sqrt{n}$. Then $N = 99\sqrt{10} + A$ and $D = 99\sqrt{10} - A$. $S = \frac{N}{D} = \frac{99\sqrt{10} + A}{99\sqrt{10} - A}$. We need to find $\frac{1}{S} - S = \frac{D}{N} - \frac{N}{D}$. If we assume that the problem is constructed such that $S=-2$, then $\frac{1}{S} - S = \frac{1}{-2} - (-2) = -1/2 + 2 = 3/2$. If $S=-2$, then $\frac{N}{D} = -2$, which means $N = -2D$. Substituting the expressions for $N$ and $D$: $99\sqrt{10} + A = -2(99\sqrt{10} - A)$ $99\sqrt{10} + A = -198\sqrt{10} + 2A$ $99\sqrt{10} + 198\sqrt{10} = 2A - A$ $297\sqrt{10} = A$. So, if $\sum_{n=1}^{99} \sqrt{n} = 297\sqrt{10}$, then $S=-2$ and the value of $\frac{1}{S} - S$ is $3/2$. Assuming this equality holds in the context of the problem, the value is $3/2$.