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

*print*Print*list*Cite

### 3 Answers

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

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

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**