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Evaluate this sum over sum

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anemone

MHB POTW Director
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Feb 14, 2012
3,676
Simplify \(\displaystyle \frac{\sum\limits_{k=1}^{99}\sqrt{10+\sqrt{k}}}{ \sum\limits_{k=1}^{99}\sqrt{10-\sqrt{k}}}\)
 

jacks

Well-known member
Apr 5, 2012
226
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 $\displaystyle \frac{A_{n}}{B_{n}} = 1+\sqrt{2}$
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,676
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:

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

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:

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

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

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

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

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

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