Question

Find: \(\sum_{k=1}^{2007} \sqrt{1 + \frac{1}{k^2} + \frac{1}{(k+1)^2}}\)

• A. \(2008 - \frac{1}{2008}\)
• B. \(2007 - \frac{1}{2007}\)
• C. \(2007 - \frac{1}{2008}\)
• D. \(2008 - \frac{1}{2007}\)
• E. \(2008 - \frac{1}{2009}\)

Hint:

Look for telescoping patterns by factoring perfect squares inside roots.

Answer 1

The Correct Option is A

Inside root: \[ 1 + \frac{1}{k^2} + \frac{1}{(k+1)^2} = \frac{k^2(k+1)^2 + (k+1)^2 + k^2}{k^2(k+1)^2} = \frac{(k^2+k+1)^2}{k^2(k+1)^2} \] Root simplifies to: \(\frac{k^2 + k + 1}{k(k+1)} = 1 + \frac{1}{k} - \frac{1}{k+1}\). Sum telescopes to \(2008 - \frac{1}{2008}\).