# Calculate the value of the sum 1/(sqrt1+sqrt2) + 1/(sqrt2+sqrt3) + 1/(sqrt3+sqrt4) + ...+1/(sqrt99+sqrt100)

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

1/(sqrt1+ sqrt2) + 1/(sqrt2 + sqrt3) + ....+ 1/(sqrt99+ sqrt100)

We know that:

1/(sqrta + sqrt(1+a) = sqrt(a+1) -sqrta

==>sqrt2 - sqrt1 + sqrt3 -sqrt2 + ...+ sqrt100 -sqrt99

= -sqrt1 + sqrt100

= -1 + 10

= 9

neela | High School Teacher | (Level 3) Valedictorian

Posted on

1/(sqrtx +sqrt(x+1) =  [sqrt(x+1)-sqrtx]/ {(sqrt(x+1)-sqrtx))(sqrtx+sqrt(x+1)},as the numerator and denominator are multiplied by the same  sqrt x+sqrt(x+1), the adjoint of sqrtx - sqrt(x+1).

= {sqrt(x+1)-sqrtx}/ (x+1 -x)

= sqrt(x+1) - sqrtx.

Therfore the given sum could rewritten  term by term as:

sqrt(2-sqrt1)+(sqrt3-sqrt2)+(sqrt4-sqrt3)+.... (sqt101-sqrt100)

sqrt101 -sqt1

= 101^(1/2) -1

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

Because we're not allowed to keep square roots at denominator, we’ll eliminate the them from the denominator, in this manner, by multiplying and dividing with the conjugated expression:

{1/[sqrt n + sqrt (n+1)]}*{[sqrt(n+1)-sqrt n]/ [sqrt(n+1)-sqrt n]}

After simplifying like terms:

1/[sqrt n + sqrt (n+1)]=[sqrt(n+1)-sqrt n]

So that,

S=1/ (sqrt 1+sqrt 2)]+…+[1/(sqrt99 + sqrt 100)]

S=sqrt2-sqrt1+sqrt3-sqrt2+sqrt4-sqrt3+…+sqrt100-sqrt99

S=sqrt100-sqrt1=10-1=9

S=9