If what you wrote is what I read, then yes, I am absolutely sure.
$\displaystyle
\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{n-1} + \sqrt{n}} + \frac{1}{\sqrt{n} + \sqrt{n+1}} =
$
$\displaystyle
= \sum_{k=1}^n \frac{1}{\sqrt{k+1} + \sqrt{k}} = \sum_{k=1}^n \frac{1}{\sqrt{k+1} + \sqrt{k}} \frac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k+1} - \sqrt{k}} =
$
$\displaystyle
= \sum_{k=1}^n (\sqrt{k+1}-\sqrt{k}) = \sqrt{n+1}-1
$
Maybe this is what you meant?
Dear Wonderboy1953,
Well I don't understand. If you want $\displaystyle \frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{n-1} + \sqrt{n}} + \frac{1}{\sqrt{n} + \sqrt{n+1}} =\sqrt{n+1}-1=499$
Then $\displaystyle \sqrt{n+1}=500\Rightarrow{n=249999}$
If you say that you should have picked n=528 then,
$\displaystyle \frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{n-1} + \sqrt{n}} + \frac{1}{\sqrt{n} + \sqrt{n+1}} =\sqrt{528+1}-1=22$
Please clarify!!
May I ask, Wboy, why does "posting it right" seem unimportant to you?
Anyway, answer to your "CORRECT" posting above is YES!!
Would be better if this way:
1 / [sqrt(0) + sqrt(1)] + 1 / [sqrt(1) + sqrt(2)] + ..... + 1 / [sqrt(483) + sqrt(484)] = 22
Get my drift? Answer is the square root of last number (484 in this case)
In other words:
1 / [sqrt(0) + sqrt(1)] + 1 / [sqrt(1) + sqrt(2)] + ..... + 1 / [sqrt(n-1) + sqrt(n)] = sqrt(n)
Now I kinda understand why it's GREAT (when correct, of course!)