+218 Compute the sum \(\frac{1}{\sqrt{36} + \sqrt{39}} + \frac{1}{\sqrt{39} + \sqrt{42}} + \frac{1}{\sqrt{42} +\sqrt{45}} + \dots + \frac{1}{\sqrt{96} + \sqrt{99}}.\)
+128707 Multiplying each fraction by its conjugate in the numerator and denominator we get
[ sqrt 36 - sqrt 39 + sqrt 39 - sqrt 42 + sqrt42 -sqrt 45 + .....+ sqrt 93 - sqrt 96 + sqrt 96 - sqrt 99 ] / -3 =
{ All terms "cancel" except those in red } =
[ sqrt 36 - sqrt 99 ] / -3 =
[sqrt 99 - sqrt 36 ] / 3 =
[ 3sqrt 11 - 6 ] / 3 =
sqrt 11 - 2
