CAT 2008 — QA Question 18

Consider only first term,

$\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}$ = $\sqrt{1+1+\frac{1}{4}}$ = $\sqrt{\frac{9}{4}}$ = $\frac{3}{2}$ = $2-\frac{1}{2}$

Now consider fisrt two terms,

$\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}$ + $\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}$ = $\frac{3}{2}$ + $\sqrt{1+\frac{1}{4}+\frac{1}{9}}$ = $\frac{3}{2}$ + $\sqrt{\frac{49}{36}}$ = $\frac{3}{2}+\frac{7}{6}$ = $\frac{8}{3}$ = $3-\frac{1}{3}$

Similarly,

$\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}$ + $\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}$ + ... + $\sqrt{1+\frac{1}{{2007}^2}+\frac{1}{{2008}^2}}$ = $2008-\frac{1}{2008}$

Hence, option (a).