SSC CGL 30 Aug 2015 Paper

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Question : 108 of 200

Marks: +1, -0

The value of

1/{1+√2}

+

1/{√2+√3}

+

1/{√3+√4}

+ ... +

1/{√8+√9}

is

1
0
2
$√2$

Solution:

1/{1+√2}

=

(√2-1)/{(√2+1)(√2-1)}

(Rationalising the denominator)
=

{√2-1}/{2-1}

=

√2

- 1
Similarly,

1/{√2+√3}

=

√3-√2

;

1/{√4+√3}

=

√4-√3

... ;

1/{√8+√9}

=

√9-√8

∴ Expression
=

√2

- 1 +

√3-√2+√4

-

√3+√5-√4

+

√6-√5

+

√7-√5+√8

-

√7+√9

-

√8

=

√9

- 1 = 3 - 1 = 2

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