1. ## Simplifying an Expression with Radicals

Can any of you simplify this expression?

Thank you!

2. ## Re: Simplifying an Expression with Radicals

Yes, by inspection it's $\sqrt{2}+1$

3. ## Re: Simplifying an Expression with Radicals

Originally Posted by jigogwapo16
Can any of you simplify this expression?
Thank you!

Real & imaginary parts of the ratio for
n=1,2,3,....,100 :: Real
n=101,102,...,300 :: Imaginary

4. ## Re: Simplifying an Expression with Radicals

How do I solve this analytically using algebra? Or would it be impossible?

Thank you!

5. ## Re: Simplifying an Expression with Radicals

Sorry for the delay, I had to ask my girlfriend how to do this...

We are given:

$\frac{\sum_{k=1}^{99}(\sqrt{10+\sqrt{k}})}{\sum_{k =1}^{99}(\sqrt{10-\sqrt{k}})}$

Let:

$r=\sqrt{10+\sqrt{k}}-\sqrt{10-\sqrt{k}}$

$r^2=10+\sqrt{k}-2\sqrt{100-k}+10-\sqrt{k}$

$r^2=2(10-\sqrt{100-k})$

Since $0 we may write:

$r=\sqrt{2}\sqrt{10-\sqrt{100-k}}$

Hence, we may write:

$\frac{\sum_{k=1}^{99}(\sqrt{10+\sqrt{k}})}{\sum_{k =1}^{99}(\sqrt{10-\sqrt{k}})}=$

$\frac{\sum_{k=1}^{99}(r+\sqrt{10-\sqrt{k}})}{\sum_{k=1}^{99}(\sqrt{10-\sqrt{k}})}=$

$\frac{\sum_{k=1}^{99}(r)+\sum_{k=1}^{99}(\sqrt{10-\sqrt{k}})}{\sum_{k=1}^{99}(\sqrt{10-\sqrt{k}})}=$

$\frac{\sum_{k=1}^{99}(r)}{\sum_{k=1}^{99}(\sqrt{10-\sqrt{k}})}+\frac{\sum_{k=1}^{99}(\sqrt{10-\sqrt{k}})}{\sum_{k=1}^{99}(\sqrt{10-\sqrt{k}})}=$

$\frac{\sum_{k=1}^{99}(\sqrt{2}\sqrt{10-\sqrt{100-k}})}{\sum_{k=1}^{99}(\sqrt{10-\sqrt{k}})}+1=$

$\sqrt{2}\frac{\sum_{k=1}^{99}(\sqrt{10-\sqrt{k}})}{\sum_{k=1}^{99}(\sqrt{10-\sqrt{k}})}+1=$

$\sqrt{2}+1$

6. ## Re: Simplifying an Expression with Radicals

^^^^hhhooolllyyycccrrraaappp.... mind... blown...
Also, Do you think they know summation notation?

7. ## Re: Simplifying an Expression with Radicals

Given the expression the OP was asked to simplify, I would assume knowledge of summation or sigma notation.

Very true.

9. ## Re: Simplifying an Expression with Radicals

Confused me...