MHB Can you generalize the result for this sum over sum problem?

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The discussion focuses on simplifying the expression $$\frac{\sum_{k=1}^{99}\sqrt{10+\sqrt{k}}}{\sum_{k=1}^{99}\sqrt{10-\sqrt{k}}$$ and generalizing it to $$\frac{\sum_{k=1}^{n^2-1}\sqrt{n+\sqrt{k}}}{\sum_{k=1}^{n^2-1}\sqrt{n-\sqrt{k}}$$ for any integer n greater than 1. The solution involves defining two sums, A_n and B_n, and deriving a relationship between them, ultimately showing that A_n equals B_n multiplied by (1 + √2). The final result indicates that the ratio of these sums simplifies to 1 + √2. The discussion concludes with a positive acknowledgment of the solution's clarity and correctness.
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Simplify $$\frac{\sum\limits_{k=1}^{99}\sqrt{10+\sqrt{k}}}{ \sum\limits_{k=1}^{99}\sqrt{10-\sqrt{k}}}$$
 
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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 $\displaystyle \frac{A_{n}}{B_{n}} = 1+\sqrt{2}$
 
jacks said:
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 $\displaystyle \frac{A_{n}}{B_{n}} = 1+\sqrt{2}$

Hi jacks,

Thanks for participating and hey, you're a "new blood" to chime in my challenge problems and welcome to the challenge problem forum!(Sun)

Yes, your solution is correct, neatly written and easy to follow, well done!

Here is the way I solve the problem:

I too generalized to compute the value of the expression:

$$ \frac{\sum\limits_{k=1}^{n^2-1}\sqrt{n+\sqrt{k}}}{ \sum\limits_{k=1}^{n^2-1}\sqrt{n-\sqrt{k}}}$$

Let:

$$r=\sqrt{n+\sqrt{k}}-\sqrt{n-\sqrt{k}}$$

$$r^2=n+\sqrt{k}-2\sqrt{n^2-k}+n-\sqrt{k}$$

$$r^2=2\left(n^2-\sqrt{n^2-k} \right)$$

Since $0<r$, we may write:

$$r=\sqrt{2}\sqrt{n^2-\sqrt{n^2-k}}$$

Hence, we may rewrite the given expression as:

$$\frac{ \sum \limits_{k=1}^{n^2-1} \sqrt{n+ \sqrt{k}}}{ \sum \limits_{k=1}^{n^2-1} \sqrt{n- \sqrt{k}}}= \frac{ \sum \limits_{k=1}^{n^2-1} \left(r+ \sqrt{n- \sqrt{k}} \right)}{ \sum \limits_{k=1}^{n^2-1} \sqrt{n- \sqrt{k}}}$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{ \sum \limits_{k=1}^{n^2-1}(r)+ \sum \limits_{k=1}^{n^2-1} \sqrt{n- \sqrt{k}}}{ \sum \limits_{k=1}^{n^2-1} \sqrt{n- \sqrt{k}}}$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{ \sum \limits_{k=1}^{n^2-1}(r)}{ \sum \limits_{k=1}^{n^2-1} \sqrt{n- \sqrt{k}}}+ \frac{ \sum \limits_{k=1}^{n^2-1} \sqrt{n- \sqrt{k}}}{ \sum \limits_{k=1}^{n^2-1} \sqrt{n- \sqrt{k}}}$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{ \sum \limits_{k=1}^{n^2-1} \left( \sqrt{2} \sqrt{n- \sqrt{n^2-k}} \right)}{ \sum \limits_{k=1}^{n^2-1} \sqrt{n- \sqrt{k}}}+1$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \sqrt{2} \frac{ \sum \limits_{k=1}^{n^2-1} \sqrt{n- \sqrt{k}}}{ \sum \limits_{k=1}^{n^2-1} \sqrt{n- \sqrt{k}}}+1$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \sqrt{2}+1$$
 
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