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

[anemone]

Simplify \(\displaystyle \frac{\sum\limits_{k=1}^{99}\sqrt{10+\sqrt{k}}}{ \sum\limits_{k=1}^{99}\sqrt{10-\sqrt{k}}}\)

 

[juantheron]

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 $\displaystyle \frac{A_{n}}{B_{n}} = 1+\sqrt{2}$

 

[anemone]

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:

Spoiler

I too generalized to compute the value of the expression:

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

Let:

\(\displaystyle r=\sqrt{n+\sqrt{k}}-\sqrt{n-\sqrt{k}}\)

\(\displaystyle r^2=n+\sqrt{k}-2\sqrt{n^2-k}+n-\sqrt{k}\)

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

Since $0<r$, we may write:

\(\displaystyle r=\sqrt{2}\sqrt{n^2-\sqrt{n^2-k}}\)

Hence, we may rewrite the given expression as:

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

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

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

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

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

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \sqrt{2}+1\)