Find the value of

Posted July 23rd, 2008 by alpha

Find the value of :

$\sqrt{1+\frac{1}{1^2 }+\frac{1}{2^2 } }+\sqrt{1+\frac{1}{2^2 }+\frac{1}{3^2 } }+............\sqrt{1+\frac{1}{999^2}+\frac{1}{1000^2 } }$

I am unable to identify the sequence.This does not seem to be an A.P.

Please show all the steps.

Comments

On July 24th, 2008 Structure says:

The only important fact is

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

Your problem

$s_{999}=\sum_{k=1}^{k=999}\sqrt{1+\frac{1}{k^2}+\frac{1}{(k+1)^2}}=999+1-\frac{1}{1000}=1000-\frac{1}{1000}=999.999$