1/sqrt(2 + sqrt(3)) + 1/sqrt(3 + sqrt(4)) + .... + 1/sqrt(1000000 + sqrt(1000001)) = sqrt( a - sqrt(b)).
a + b = .... ?

Question

1/sqrt(2 + sqrt(3)) + 1/sqrt(3 + sqrt(4)) + .... + 1/sqrt(1000000 + sqrt(1000001)) = sqrt( a - sqrt(b)).
a + b = .... ?

anonymous · Answer

@ganeshie8 and @dan815 
help me ... how to simplify this summation of radicals :D

inkyvoyd · Answer

rationalize the denominator... some stuff should cancel out :)

kainui · Answer

Totally do what inky says, I just thought I'd share some stuff anyways, kinda incomplete but if you're curious I can explain more. My systematic way of solving these lately has been to write it in summation notation:

$$\sum_{n=a}^b g(n)$$

then look for solutions to:

$$g(n) = f(n+1)-f(n)$$ or $$g(n) = f(n)-f(n-1)$$

The point is, then your sum becomes a telescoping series so you only have to evaluate the end points. Really when written out, it's basically the fundamental theorem of calculus:

$$\sum_{n=a}^{b-1} \Delta f(n) = f(b)-f(a)$$

ganeshie8 · Answer

rationalizing doesn't help much here

wcrmelissa2001 · Answer

not sure how to solve this but I think normally how it works for these kind of questions is that you simplify/rationalise/something the first one or two numbers and look for a pattern. This isn't very helpful sorry but I did a similar question before and found that each number actually was 1......yea....

kainui · Answer

@tanjung Are you sure you wrote the question correctly?

anonymous · Answer

Yes. That is right equations...

anonymous · Answer

Sorry i cant write it in latex. I want show you in picture,  but i cant do it by using android

kainui · Answer

$$\sum_{n=2}^{10^6} \frac{1}{\sqrt{n+\sqrt{n+1}}} = \sqrt{a - \sqrt{b}}$$

What is $a+b$

anonymous · Answer

Yep. That is like above. Thanks :D

kainui · Answer

How'd you get this question this seems tough lol

anonymous · Answer

My friend 's question.. i tried help her, but stuck also :v