+0  
 
+1
3
1
avatar+218 

Compute the sum \(\frac{1}{\sqrt{36} + \sqrt{39}} + \frac{1}{\sqrt{39} + \sqrt{42}} + \frac{1}{\sqrt{42} +\sqrt{45}} + \dots + \frac{1}{\sqrt{96} + \sqrt{99}}.\)

 Apr 29, 2024
 #1
avatar+128772 
+1

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

 

cool cool cool

 Apr 29, 2024

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