Some easy sequences

rationalise the denominator. On rationalising you will find that all terms cancel off except sq.rt(1) and sq.rt(6400). So on adding we find that n^2 =81 mod(n)=9

$\frac { 1 }{ \sqrt { 2 } -\sqrt { 1 } } -\frac { 1 }{ \sqrt { 3 } -\sqrt { 2 } } +\frac { 1 }{ \sqrt { 4 } -\sqrt { 3 } } ...+\frac { 1 }{ \sqrt { 6400 } -\sqrt { 6399 } } =$

$\frac { \sqrt { 2 } +\sqrt { 1 } }{ 2-1 } -\frac { \sqrt { 3 } +\sqrt { 2 } }{ 3-2 } +\frac { \sqrt { 4 } +\sqrt { 3 } }{ 4-3 } ...+\frac { \sqrt { 6400 } +\sqrt { 6399 } }{ 6400-6399 }$

Now that we have a common denominator of $1$ , we can ignore the denominator. Notice that all the terms cancel each other out except for the two end terms, $\sqrt { 6400 }$ and $1$ . Thus,

${ n }^{ 2 }=81\quad \Rightarrow \quad \left| n \right| =9$