What is the simplest form of the sum 1/(1+square root 2) + 1/(square root 2+square root 3)+...........+1/(square root 2000+square root2001)?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to simplify 1/(1+ sqrt 2) + 1/(sqrt 2 + sqrt 3) +...........+ 1/(sqrt 2000 + sqrt 2001)

multiply the numrator and denominator of each term of the form 1/(sqrt a + sqrt b) by (sqrt a - sqrt b)

=> (sqrt a - sqrt b)/( a - b)

1/(1+ sqrt 2) + 1/(sqrt 2 + sqrt 3) +...........+ 1/(sqrt 2000 + sqrt 2001)

=> (1 - sqrt 2)/(-1) + (sqrt 2 - sqrt 3)/(-1) + ... + (sqrt 2000 - sqrt 2001)/(-1)

=> (-1)[ 1 - sqrt 2 + sqrt 2 - sqrt 3 + sqrt 3 +...+sqrt 2000 - sqrt 2001)

=> -1 * ( 1 - sqrt 2001)

=> sqrt 2001 - 1

The required simplified form is sqrt 2001 - 1

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll write the general term of the sum:

1/(sqrt k + sqrt(k+1))

We'll multiply the numerator and denominator by the conjugate of denominator:

(sqrt k - sqrt(k+1))/[(sqrt k)^2 - (sqrt(k+1))^2] = (sqrt k - sqrt(k+1))/(k - k - 1)

1/(sqrt k + sqrt(k+1)) = (sqrt k - sqrt(k+1))/-1

1/(sqrt k + sqrt(k+1)) = sqrt(k+1) - sqrt k

We'll put k=1 => 1/(1+sqrt2) = sqrt 2 - 1

We'll put k = 2=> 1/(sqrt 2+sqrt 3) = sqrt 3 - sqrt 2

.................................................................................

We'll put k = 2000=> 1/(sqrt 2000+sqrt 2001) = sqrt 2001 - sqrt 2000

We'll add the terms:

1/(1+sqrt2) + ... + 1/(sqrt 2000+sqrt 2001) = sqrt 2 - 1 + sqrt 3 - sqrt 2 + ... + sqrt 2001 - sqrt 2000

We'll eliminate like terms:

1/(1+sqrt2) + ... + 1/(sqrt 2000+sqrt 2001) = sqrt 2001 - 1

The simplest form of the given sum is: S = sqrt 2001 - 1.

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