Some easy sequences

Algebra Level pending

1 2 1 1 3 2 + 1 4 3 1 5 4 + + 1 6400 6399 = n 2 \frac { 1 }{ \sqrt { 2 } -\sqrt { 1 } } -\frac { 1 }{ \sqrt { 3 } -\sqrt { 2 } } +\frac { 1 }{ \sqrt { 4 } -\sqrt { 3 } } -\frac { 1 }{ \sqrt { 5 } -\sqrt { 4 } } + \cdots +\frac { 1 }{ \sqrt { 6400 } -\sqrt { 6399 } } = { n }^{ 2 }

Determine n \left| n \right| .


The answer is 9.

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2 solutions

Harshita Moondra
Apr 22, 2015

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

Dylan Pentland
Mar 26, 2015

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

2 + 1 2 1 3 + 2 3 2 + 4 + 3 4 3 . . . + 6400 + 6399 6400 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 1 , we can ignore the denominator. Notice that all the terms cancel each other out except for the two end terms, 6400 \sqrt { 6400 } and 1 1 . Thus,

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

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