Question

If, $\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+.....+\sqrt{1+\frac{1}{(1999{)}^2}+\frac{1}{(2000{)}^2}}=x-\frac{1}{x}$, then find the value of $x$.

Answer 1

$\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+.....+\sqrt{1+\frac{1}{(1999{)}^2}+\frac{1}{(2000{)}^2}}$

$=x-\frac{1}{x}$

Considering each term,

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

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

$\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}=\sqrt{1+\frac{1}{9}+\frac{1}{16}}=\sqrt{\frac{169}{144}}=\frac{13}{12}=1+\frac{1}{12}$

So,

$=1+\frac{1}{2}+1+\frac{1}{6}+1+\frac{1}{12}+......+1+\frac{1}{\left(1999\right)\left(2000\right)}$

$=1999+[\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+......+\frac{1}{\left(1999\right)\left(2000\right)}]$

$=1999+[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{1999}-\frac{1}{2000}]$

$=1999+1-\frac{1}{2000}=2000-\frac{1}{2000}$

So, $x=2000$ (Answer)