calculate the value :
$$\sqrt{1+\frac{ 1 }{ 1^2 }+\frac{ 1 }{ 2^2 }} + \sqrt{1+\frac{ 1 }{ 2^2 }+\frac{ 1 }{ 3^2 }}+\sqrt{1+\frac{ 1 }{ 3^2 }+\frac{ 1 }{ 4^2 }}+...+\sqrt{1+\frac{ 1 }{ 2004^2 }+\frac{ 1 }{ 2005^2 }}$$

Question

calculate the value :
$$\sqrt{1+\frac{ 1 }{ 1^2 }+\frac{ 1 }{ 2^2 }} + \sqrt{1+\frac{ 1 }{ 2^2 }+\frac{ 1 }{ 3^2 }}+\sqrt{1+\frac{ 1 }{ 3^2 }+\frac{ 1 }{ 4^2 }}+...+\sqrt{1+\frac{ 1 }{ 2004^2 }+\frac{ 1 }{ 2005^2 }}$$

anonymous · Answer

That's one sweet sum :P
I'm doing.

anonymous · Answer

Now I have to start over, stupid crash

freckles · Answer

I was able to find the sum by finding a pattern in the first few sums

anonymous · Answer

Nope, I got the eqation wrong again, darn

freckles · Answer

that 
is from 1 to 2 we have 8/3
and from 1 to 3 we have 15/4
and from 1 to 4 we have 24/5

and so on...

anonymous · Answer

the trouble with this sum is , it is under roots
I'm still thinking what to do

freckles · Answer

$$\sum_{i=1}^{n}\sqrt{1+\frac{1}{x^2}+\frac{1}{(x+1)^2}}=\frac{n(n+2)}{n+1}$$

freckles · Answer

that is the pattern I observe

freckles · Answer

from the first few sums

rational · Answer

$$\large \sqrt{1+\dfrac{1}{n^2} +\frac{1}{(n+1)^2}} = \sqrt{\dfrac{(n^2+n+1)^2}{n^2(n+1)^2}}$$

freckles · Answer

also replace x with i *

freckles · Answer

that was a type-o

anonymous · Answer

@rational I dont think that equation would solve to the answer.

rational · Answer

cancel the radical and see if it telescopes

freckles · Answer

I like @rational 's way more than mine
i like telescoping thingys

anonymous · Answer

@chihiroasleaf  Tel me this, is it equated to anything on the left hand side or purely asked for a left hand side value?

chihiroasleaf · Answer

thanks a bunch @rational  :)
so, 

$$\sqrt{1+\frac{ 1 }{ n^2 }+\frac{ 1 }{ (n+1)^2 }} = 1 + \frac{ 1 }{ n(n+1) }$$

therefore:
$$\sqrt{1+\frac{ 1 }{ 1^2 }+\frac{ 1 }{ 2^2 }} + \sqrt{1+\frac{ 1 }{ 2^2 }+\frac{ 1 }{ 3^2 }}+\sqrt{1+\frac{ 1 }{ 3^2 }+\frac{ 1 }{ 4^2 }}+...+\sqrt{1+\frac{ 1 }{ 2004^2 }+\frac{ 1 }{ 2005^2 }}$$
$$=1+\frac{ 1 }{ 1 \times 2 }+1+ \frac{ 1 }{ 2 \times 3 }+.....+1+\frac{ 1 }{2004 \times 2005 }$$
$$=2004+\frac{ 1 }{ 1 }-\frac{ 1 }{ 2 }+\frac{ 1 }{ 2 }-\frac{ 1 }{ 3 }+....+\frac{ 1 }{ 2004 }-\frac{ 1 }{ 2005 }$$
$$=2004+\frac{ 2004 }{ 2005 } =2004\frac{ 2004 }{ 2005} $$

chihiroasleaf · Answer

@srinuscooby  it's purely ask the value of the expression

rational · Answer

Looks neat !!

rational · Answer

In general : 

$$\large \sum\limits_{r=1}^{n} \left[f(r) - f(r+1) \right]= f(1) - f(n+1) $$