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Summation of series

  1. Jan 16, 2014 #1
    1. The problem statement, all variables and given/known data
    Someone please check my work... :D

    If ##f(x)=\sqrt{x}+\sqrt{x+1}## , find the value of
    ##\frac{1}{f(1)}+\frac{1}{f(2)}+\frac{1}{f(3)}+...+\frac{1}{f(24)}##

    2. Relevant equations
    Summation of series, rationalizing the denominator.


    3. The attempt at a solution
    ##f(x)=\sqrt{x}+\sqrt{x+1}##
    ##f(x)=\sqrt{x}+\sqrt{x+1}(\frac{\sqrt{x}-\sqrt{x+1}}{\sqrt{x}-\sqrt{x+1}})##
    ##f(x)=\frac{x-(x+1)}{\sqrt{x}-\sqrt{x+1}}##
    ##\frac{1}{f(x)}=\sqrt{x+1}-\sqrt{x}##
    ##\frac{1}{f(1)}+\frac{1}{f(2)}+\frac{1}{f(3)}+...+\frac{1}{f(24)}##
    ##=\sum_{r=1}^{24} \sqrt{x+1}-\sqrt{x}##
    ##=(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+...+(\sqrt{24}-\sqrt{23})+(\sqrt{25}-\sqrt{24})##
    ##=\sqrt{25}-\sqrt{1}##
    ##=4##
     
    Last edited by a moderator: Jan 16, 2014
  2. jcsd
  3. Jan 16, 2014 #2

    vanhees71

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    Science Advisor
    2016 Award

    I would have done it a bit shorter
    [tex]\frac{1}{f(x)}=\frac{1}{\sqrt{x+1}+\sqrt{x}}=\frac{\sqrt{x+1}-\sqrt{x}}{(\sqrt{x+1}-\sqrt{x})(\sqrt{x+1}+\sqrt{x})}=\sqrt{x+1}-\sqrt{x}.[/tex]
    Your result is nevertheless correct.
     
  4. Jan 16, 2014 #3
    Thanks a lot! :D
     
  5. Jan 16, 2014 #4

    Borek

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    Staff: Mentor

    Your parentheses were slightly off.
     
  6. Jan 16, 2014 #5
    Edited.
     
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