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Simplify  \(\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \dots + \frac{1}{\sqrt{99} + \sqrt{100}}.\)

 Jun 27, 2019

Best Answer 

 #2
avatar+8963 
+5

\(S\ =\ \frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \dots + \frac{1}{\sqrt{99} + \sqrt{100}}\\~\\ S\ =\ \sum\limits_{n=1}^{99}\frac{1}{\sqrt{n}+\sqrt{n+1}}\\~\\ S\ =\ \sum\limits_{n=1}^{99}\frac{1}{\sqrt{n}+\sqrt{n+1}}\cdot \frac{\sqrt{n}-\sqrt{n+1}}{\sqrt{n}-\sqrt{n+1}}\\~\\ S\ =\ \sum\limits_{n=1}^{99}\frac{\sqrt{n}-\sqrt{n+1}}{(n)-(n+1)}\\~\\ S\ =\ \sum\limits_{n=1}^{99}\frac{\sqrt{n}-\sqrt{n+1}}{-1}\\~\\ S\ =\ \sum\limits_{n=1}^{99}\sqrt{n+1}-\sqrt{n}\\~\\ S\ =\ {\color{}\sqrt2}-\sqrt1\ {\color{}+\sqrt3-\sqrt2+\sqrt4-\sqrt3+\sqrt5-\sqrt4+\dots+\sqrt{99}-\sqrt{98}}+\sqrt{100}\ {\color{}-\ \sqrt{99}}\\~\\ S\ =\ {\color{gray}\sqrt2}-\sqrt1\ {\color{gray}+\sqrt3-\sqrt2+\sqrt4-\sqrt3+\sqrt5-\sqrt4+\dots+\sqrt{99}-\sqrt{98}}+\sqrt{100}\ {\color{gray}-\ \sqrt{99}}\\~\\ S\ =\ -\sqrt1+\sqrt{100}\\~\\ S\ =\ -1+10\\~\\ S\ =\ 9 \)_

 Jun 27, 2019
 #1
avatar+111396 
+2

   1 (1 - √2)                  (1  - √2)             

______________   =___________  =     ( √2 - 1)

(1 + √2) ( 1 - √2)               -1

 

   1 ( √2 - √3)                  (√2  - √3)             

______________   =___________  =     ( √3 - √2)

(√2+ √3) ( √2 - √3)               -1

 

 

   1 ( √3 - √4)                   (√3  - √4)             

_______________   =  ___________  =     ( √4 - √3)

(√3 + √4) ( √3 - √4)               -1

 

So we have

 

-1  +   ( √2 - √2) + ( √3 - √3)  + ....... +  ( √99 - √99)  +    √100     =

 

-1   +         0       +         0        +........+          0         +     √100     =

 

-1  + √100 =

 

-1 +  10 =

 

9

 

 

cool cool cool

 Jun 27, 2019
 #2
avatar+8963 
+5
Best Answer

\(S\ =\ \frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \dots + \frac{1}{\sqrt{99} + \sqrt{100}}\\~\\ S\ =\ \sum\limits_{n=1}^{99}\frac{1}{\sqrt{n}+\sqrt{n+1}}\\~\\ S\ =\ \sum\limits_{n=1}^{99}\frac{1}{\sqrt{n}+\sqrt{n+1}}\cdot \frac{\sqrt{n}-\sqrt{n+1}}{\sqrt{n}-\sqrt{n+1}}\\~\\ S\ =\ \sum\limits_{n=1}^{99}\frac{\sqrt{n}-\sqrt{n+1}}{(n)-(n+1)}\\~\\ S\ =\ \sum\limits_{n=1}^{99}\frac{\sqrt{n}-\sqrt{n+1}}{-1}\\~\\ S\ =\ \sum\limits_{n=1}^{99}\sqrt{n+1}-\sqrt{n}\\~\\ S\ =\ {\color{}\sqrt2}-\sqrt1\ {\color{}+\sqrt3-\sqrt2+\sqrt4-\sqrt3+\sqrt5-\sqrt4+\dots+\sqrt{99}-\sqrt{98}}+\sqrt{100}\ {\color{}-\ \sqrt{99}}\\~\\ S\ =\ {\color{gray}\sqrt2}-\sqrt1\ {\color{gray}+\sqrt3-\sqrt2+\sqrt4-\sqrt3+\sqrt5-\sqrt4+\dots+\sqrt{99}-\sqrt{98}}+\sqrt{100}\ {\color{gray}-\ \sqrt{99}}\\~\\ S\ =\ -\sqrt1+\sqrt{100}\\~\\ S\ =\ -1+10\\~\\ S\ =\ 9 \)_

hectictar Jun 27, 2019
 #3
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0

This sequence has this "closed form":

Sum =Sqrt(n) - 1, where n = Last term

Sum =Sqrt(100) - 1 = 9

IF: n = 4, Sum =Sqrt(4) - 1 = 1

      n = 121, Sum = Sqrt(121) - 1 = 10

      n = 961, Sum = Sqrt(961) - 1 = 30........and so on.

 Jun 27, 2019

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