Using mathematical induction to solve a problem












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I was solving some mathematical induction problems when I ran into this one and couldn't figure out how to solve it.



Would someone be able to give me a breakdown on how to complete it?



$$frac23+frac29+frac2{27}+...+frac{2}{3^n}=1-frac1{3^n}$$






discrete-mathematics induction







share|cite|improve this question









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  • $begingroup$
    As an aside, note that this is true for the same reason as $0.999 = 1-frac{1}{1000}$.
    $endgroup$
    – Patrick Stevens
    Dec 18 '18 at 21:41










  • $begingroup$
    The sum on left is a (finite) geometric series, ratio $1/3.$
    $endgroup$
    – coffeemath
    Dec 18 '18 at 21:45
















0












$begingroup$


I was solving some mathematical induction problems when I ran into this one and couldn't figure out how to solve it.



Would someone be able to give me a breakdown on how to complete it?



$$frac23+frac29+frac2{27}+...+frac{2}{3^n}=1-frac1{3^n}$$






discrete-mathematics induction







share|cite|improve this question









$endgroup$












  • $begingroup$
    As an aside, note that this is true for the same reason as $0.999 = 1-frac{1}{1000}$.
    $endgroup$
    – Patrick Stevens
    Dec 18 '18 at 21:41










  • $begingroup$
    The sum on left is a (finite) geometric series, ratio $1/3.$
    $endgroup$
    – coffeemath
    Dec 18 '18 at 21:45














0












0








0


0



$begingroup$


I was solving some mathematical induction problems when I ran into this one and couldn't figure out how to solve it.



Would someone be able to give me a breakdown on how to complete it?



$$frac23+frac29+frac2{27}+...+frac{2}{3^n}=1-frac1{3^n}$$






discrete-mathematics induction







share|cite|improve this question









$endgroup$




I was solving some mathematical induction problems when I ran into this one and couldn't figure out how to solve it.



Would someone be able to give me a breakdown on how to complete it?



$$frac23+frac29+frac2{27}+...+frac{2}{3^n}=1-frac1{3^n}$$






discrete-mathematics induction




discrete-mathematics induction






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share|cite|improve this question











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asked Dec 18 '18 at 21:35









bsoleimanybsoleimany

81




81












  • $begingroup$
    As an aside, note that this is true for the same reason as $0.999 = 1-frac{1}{1000}$.
    $endgroup$
    – Patrick Stevens
    Dec 18 '18 at 21:41










  • $begingroup$
    The sum on left is a (finite) geometric series, ratio $1/3.$
    $endgroup$
    – coffeemath
    Dec 18 '18 at 21:45


















  • $begingroup$
    As an aside, note that this is true for the same reason as $0.999 = 1-frac{1}{1000}$.
    $endgroup$
    – Patrick Stevens
    Dec 18 '18 at 21:41










  • $begingroup$
    The sum on left is a (finite) geometric series, ratio $1/3.$
    $endgroup$
    – coffeemath
    Dec 18 '18 at 21:45
















$begingroup$
As an aside, note that this is true for the same reason as $0.999 = 1-frac{1}{1000}$.
$endgroup$
– Patrick Stevens
Dec 18 '18 at 21:41




$begingroup$
As an aside, note that this is true for the same reason as $0.999 = 1-frac{1}{1000}$.
$endgroup$
– Patrick Stevens
Dec 18 '18 at 21:41












$begingroup$
The sum on left is a (finite) geometric series, ratio $1/3.$
$endgroup$
– coffeemath
Dec 18 '18 at 21:45




$begingroup$
The sum on left is a (finite) geometric series, ratio $1/3.$
$endgroup$
– coffeemath
Dec 18 '18 at 21:45










2 Answers
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$begingroup$

I will give you the induction step $nmapsto n+1$:



$frac{2}{3}+ldots + frac{2}{3^{n+1}} = left(1-frac{1}{3^n}right)+frac{2}{3^{n+1}} = 1 + left(frac{2}{3^{n+1}}-frac{1}{3^n}right) = 1+frac{2}{3^{n+1}}-frac{3}{3cdot 3^{n}}= 1-frac{1}{3^{n+1}}$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Since you specifically ask about induction:
    For n= 1 this is $frac{2}{3}= frac{1}{1- frac{1}{3}}$ which is true.



    Now, assume that, for some k, $frac{2}{3}+ frac{2}{9}+ cdotcdotcdot+ frac{2}{3^k}= 1- frac{1}{3^k}$. Then with $n= k+ 1$ we have $frac{2}{3}+ frac{2}{9}+ cdotcdotcdot+ frac{2}{3^k}+ frac{2}{3^{k+1}}= 1- frac{1}{3^k}+ frac{2}{3^{k+1}}= 1- frac{3}{3^{k+1}}+ frac{2}{3^{k+1}}= 1- frac{1}{3^{k+1}}$.






    share|cite|improve this answer









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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      active

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      active

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      2












      $begingroup$

      I will give you the induction step $nmapsto n+1$:



      $frac{2}{3}+ldots + frac{2}{3^{n+1}} = left(1-frac{1}{3^n}right)+frac{2}{3^{n+1}} = 1 + left(frac{2}{3^{n+1}}-frac{1}{3^n}right) = 1+frac{2}{3^{n+1}}-frac{3}{3cdot 3^{n}}= 1-frac{1}{3^{n+1}}$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        I will give you the induction step $nmapsto n+1$:



        $frac{2}{3}+ldots + frac{2}{3^{n+1}} = left(1-frac{1}{3^n}right)+frac{2}{3^{n+1}} = 1 + left(frac{2}{3^{n+1}}-frac{1}{3^n}right) = 1+frac{2}{3^{n+1}}-frac{3}{3cdot 3^{n}}= 1-frac{1}{3^{n+1}}$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          I will give you the induction step $nmapsto n+1$:



          $frac{2}{3}+ldots + frac{2}{3^{n+1}} = left(1-frac{1}{3^n}right)+frac{2}{3^{n+1}} = 1 + left(frac{2}{3^{n+1}}-frac{1}{3^n}right) = 1+frac{2}{3^{n+1}}-frac{3}{3cdot 3^{n}}= 1-frac{1}{3^{n+1}}$






          share|cite|improve this answer









          $endgroup$



          I will give you the induction step $nmapsto n+1$:



          $frac{2}{3}+ldots + frac{2}{3^{n+1}} = left(1-frac{1}{3^n}right)+frac{2}{3^{n+1}} = 1 + left(frac{2}{3^{n+1}}-frac{1}{3^n}right) = 1+frac{2}{3^{n+1}}-frac{3}{3cdot 3^{n}}= 1-frac{1}{3^{n+1}}$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 18 '18 at 21:50









          Card_TrickCard_Trick

          935




          935























              0












              $begingroup$

              Since you specifically ask about induction:
              For n= 1 this is $frac{2}{3}= frac{1}{1- frac{1}{3}}$ which is true.



              Now, assume that, for some k, $frac{2}{3}+ frac{2}{9}+ cdotcdotcdot+ frac{2}{3^k}= 1- frac{1}{3^k}$. Then with $n= k+ 1$ we have $frac{2}{3}+ frac{2}{9}+ cdotcdotcdot+ frac{2}{3^k}+ frac{2}{3^{k+1}}= 1- frac{1}{3^k}+ frac{2}{3^{k+1}}= 1- frac{3}{3^{k+1}}+ frac{2}{3^{k+1}}= 1- frac{1}{3^{k+1}}$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Since you specifically ask about induction:
                For n= 1 this is $frac{2}{3}= frac{1}{1- frac{1}{3}}$ which is true.



                Now, assume that, for some k, $frac{2}{3}+ frac{2}{9}+ cdotcdotcdot+ frac{2}{3^k}= 1- frac{1}{3^k}$. Then with $n= k+ 1$ we have $frac{2}{3}+ frac{2}{9}+ cdotcdotcdot+ frac{2}{3^k}+ frac{2}{3^{k+1}}= 1- frac{1}{3^k}+ frac{2}{3^{k+1}}= 1- frac{3}{3^{k+1}}+ frac{2}{3^{k+1}}= 1- frac{1}{3^{k+1}}$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Since you specifically ask about induction:
                  For n= 1 this is $frac{2}{3}= frac{1}{1- frac{1}{3}}$ which is true.



                  Now, assume that, for some k, $frac{2}{3}+ frac{2}{9}+ cdotcdotcdot+ frac{2}{3^k}= 1- frac{1}{3^k}$. Then with $n= k+ 1$ we have $frac{2}{3}+ frac{2}{9}+ cdotcdotcdot+ frac{2}{3^k}+ frac{2}{3^{k+1}}= 1- frac{1}{3^k}+ frac{2}{3^{k+1}}= 1- frac{3}{3^{k+1}}+ frac{2}{3^{k+1}}= 1- frac{1}{3^{k+1}}$.






                  share|cite|improve this answer









                  $endgroup$



                  Since you specifically ask about induction:
                  For n= 1 this is $frac{2}{3}= frac{1}{1- frac{1}{3}}$ which is true.



                  Now, assume that, for some k, $frac{2}{3}+ frac{2}{9}+ cdotcdotcdot+ frac{2}{3^k}= 1- frac{1}{3^k}$. Then with $n= k+ 1$ we have $frac{2}{3}+ frac{2}{9}+ cdotcdotcdot+ frac{2}{3^k}+ frac{2}{3^{k+1}}= 1- frac{1}{3^k}+ frac{2}{3^{k+1}}= 1- frac{3}{3^{k+1}}+ frac{2}{3^{k+1}}= 1- frac{1}{3^{k+1}}$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 18 '18 at 22:02









                  user247327user247327

                  11.1k1515




                  11.1k1515






























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