Proving by induction of n. Is this correct until this point?












4












$begingroup$



$$
sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac{1}{2} - frac1{(n+1)2^{n+1}}
$$




Base Case:



I did $n = 1$, so..



LHS-



$$sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac3{8}$$



RHS-



$$frac{1}{2} - frac1{(n+1)2^{n+1}} = frac3{8}$$



so LHS = RHS



Inductive case-



LHS for $n+1$



$$sum_{k=1}^{n+1} frac {k+2}{k(k+1)2^{k+1}} +frac {n+3}{(n+1)(n+2)2^{n+2}}$$



and then I think that you can use inductive hypothesis to change it to the form of
$$
frac{1}{2} - frac1{(n+1)2^{n+1}} +frac {n+3}{(n+1)(n+2)2^{n+2}}
$$



and then I broke up $frac {n+3}{(n+1)(n+2)2^{n+2}}$ into



$$frac{2(n+2)-(n+1)}{(n+1)(n+2)2^{n+2}}$$



$$=$$



$$frac{2}{(n+1)2^{n+2}} - frac{1}{(n+2)2^{n+2}}$$



$$=$$



$$frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$



then put it back in with the rest of the equation, bringing me to



$$frac{1}2 -frac {1}{(n+1)2^{n+1}} +frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$



then



$$frac{1}2 -frac{2}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$



and



$$frac{1}2 -frac{1}{(n+1)2^{n}} - frac{1}{(n+2)2^{n+2}}$$



$$frac{1}2 -frac{(n+2)2^{n+2} - (n+1)2^{n}}{(n+1)(n+2)2^{2n+2}} $$



which I think simplifies down to this after factoring out a $2^{n}$ from the numerator?



$$frac{1}2 -frac{2^{n}((n+2)2^{2} - (n+1))}{(n+1)(n+2)2^{2n+2}} $$



canceling out $2^{n}$



$$frac{1}2 -frac{(3n-7)}{(n+1)(n+2)2^{n+2}} $$



and I'm stuck, please help!






discrete-mathematics induction







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    4












    $begingroup$



    $$
    sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac{1}{2} - frac1{(n+1)2^{n+1}}
    $$




    Base Case:



    I did $n = 1$, so..



    LHS-



    $$sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac3{8}$$



    RHS-



    $$frac{1}{2} - frac1{(n+1)2^{n+1}} = frac3{8}$$



    so LHS = RHS



    Inductive case-



    LHS for $n+1$



    $$sum_{k=1}^{n+1} frac {k+2}{k(k+1)2^{k+1}} +frac {n+3}{(n+1)(n+2)2^{n+2}}$$



    and then I think that you can use inductive hypothesis to change it to the form of
    $$
    frac{1}{2} - frac1{(n+1)2^{n+1}} +frac {n+3}{(n+1)(n+2)2^{n+2}}
    $$



    and then I broke up $frac {n+3}{(n+1)(n+2)2^{n+2}}$ into



    $$frac{2(n+2)-(n+1)}{(n+1)(n+2)2^{n+2}}$$



    $$=$$



    $$frac{2}{(n+1)2^{n+2}} - frac{1}{(n+2)2^{n+2}}$$



    $$=$$



    $$frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$



    then put it back in with the rest of the equation, bringing me to



    $$frac{1}2 -frac {1}{(n+1)2^{n+1}} +frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$



    then



    $$frac{1}2 -frac{2}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$



    and



    $$frac{1}2 -frac{1}{(n+1)2^{n}} - frac{1}{(n+2)2^{n+2}}$$



    $$frac{1}2 -frac{(n+2)2^{n+2} - (n+1)2^{n}}{(n+1)(n+2)2^{2n+2}} $$



    which I think simplifies down to this after factoring out a $2^{n}$ from the numerator?



    $$frac{1}2 -frac{2^{n}((n+2)2^{2} - (n+1))}{(n+1)(n+2)2^{2n+2}} $$



    canceling out $2^{n}$



    $$frac{1}2 -frac{(3n-7)}{(n+1)(n+2)2^{n+2}} $$



    and I'm stuck, please help!






    discrete-mathematics induction







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      4












      4








      4


      1



      $begingroup$



      $$
      sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac{1}{2} - frac1{(n+1)2^{n+1}}
      $$




      Base Case:



      I did $n = 1$, so..



      LHS-



      $$sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac3{8}$$



      RHS-



      $$frac{1}{2} - frac1{(n+1)2^{n+1}} = frac3{8}$$



      so LHS = RHS



      Inductive case-



      LHS for $n+1$



      $$sum_{k=1}^{n+1} frac {k+2}{k(k+1)2^{k+1}} +frac {n+3}{(n+1)(n+2)2^{n+2}}$$



      and then I think that you can use inductive hypothesis to change it to the form of
      $$
      frac{1}{2} - frac1{(n+1)2^{n+1}} +frac {n+3}{(n+1)(n+2)2^{n+2}}
      $$



      and then I broke up $frac {n+3}{(n+1)(n+2)2^{n+2}}$ into



      $$frac{2(n+2)-(n+1)}{(n+1)(n+2)2^{n+2}}$$



      $$=$$



      $$frac{2}{(n+1)2^{n+2}} - frac{1}{(n+2)2^{n+2}}$$



      $$=$$



      $$frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$



      then put it back in with the rest of the equation, bringing me to



      $$frac{1}2 -frac {1}{(n+1)2^{n+1}} +frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$



      then



      $$frac{1}2 -frac{2}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$



      and



      $$frac{1}2 -frac{1}{(n+1)2^{n}} - frac{1}{(n+2)2^{n+2}}$$



      $$frac{1}2 -frac{(n+2)2^{n+2} - (n+1)2^{n}}{(n+1)(n+2)2^{2n+2}} $$



      which I think simplifies down to this after factoring out a $2^{n}$ from the numerator?



      $$frac{1}2 -frac{2^{n}((n+2)2^{2} - (n+1))}{(n+1)(n+2)2^{2n+2}} $$



      canceling out $2^{n}$



      $$frac{1}2 -frac{(3n-7)}{(n+1)(n+2)2^{n+2}} $$



      and I'm stuck, please help!






      discrete-mathematics induction







      share|cite|improve this question









      $endgroup$





      $$
      sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac{1}{2} - frac1{(n+1)2^{n+1}}
      $$




      Base Case:



      I did $n = 1$, so..



      LHS-



      $$sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac3{8}$$



      RHS-



      $$frac{1}{2} - frac1{(n+1)2^{n+1}} = frac3{8}$$



      so LHS = RHS



      Inductive case-



      LHS for $n+1$



      $$sum_{k=1}^{n+1} frac {k+2}{k(k+1)2^{k+1}} +frac {n+3}{(n+1)(n+2)2^{n+2}}$$



      and then I think that you can use inductive hypothesis to change it to the form of
      $$
      frac{1}{2} - frac1{(n+1)2^{n+1}} +frac {n+3}{(n+1)(n+2)2^{n+2}}
      $$



      and then I broke up $frac {n+3}{(n+1)(n+2)2^{n+2}}$ into



      $$frac{2(n+2)-(n+1)}{(n+1)(n+2)2^{n+2}}$$



      $$=$$



      $$frac{2}{(n+1)2^{n+2}} - frac{1}{(n+2)2^{n+2}}$$



      $$=$$



      $$frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$



      then put it back in with the rest of the equation, bringing me to



      $$frac{1}2 -frac {1}{(n+1)2^{n+1}} +frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$



      then



      $$frac{1}2 -frac{2}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}$$



      and



      $$frac{1}2 -frac{1}{(n+1)2^{n}} - frac{1}{(n+2)2^{n+2}}$$



      $$frac{1}2 -frac{(n+2)2^{n+2} - (n+1)2^{n}}{(n+1)(n+2)2^{2n+2}} $$



      which I think simplifies down to this after factoring out a $2^{n}$ from the numerator?



      $$frac{1}2 -frac{2^{n}((n+2)2^{2} - (n+1))}{(n+1)(n+2)2^{2n+2}} $$



      canceling out $2^{n}$



      $$frac{1}2 -frac{(3n-7)}{(n+1)(n+2)2^{n+2}} $$



      and I'm stuck, please help!






      discrete-mathematics induction




      discrete-mathematics induction






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      share|cite|improve this question











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      asked 4 hours ago









      BrownieBrownie

      1927




      1927






















          2 Answers
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          $begingroup$

          Your error is just after the sixth step from the bottom:



          $$frac{1}2 -frac {1}{(n+1)2^{n+1}} +frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}=frac{1}2 -frac{1}{(n+2)2^{n+2}}$$



          Then you are done.



          You accidentally added the two middle terms instead of subtracting.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            Using a telescoping sum, we get
            $$
            begin{align}
            sum_{k=1}^nfrac{k+2}{k(k+1)2^{k+1}}
            &=sum_{k=1}^nleft(frac1{k2^k}-frac1{(k+1)2^{k+1}}right)\
            &=sum_{k=1}^nfrac1{k2^k}-sum_{k=2}^{n+1}frac1{k2^k}\
            &=frac12-frac1{(n+1)2^{n+1}}
            end{align}
            $$






            share|cite|improve this answer









            $endgroup$













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              2 Answers
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              $begingroup$

              Your error is just after the sixth step from the bottom:



              $$frac{1}2 -frac {1}{(n+1)2^{n+1}} +frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}=frac{1}2 -frac{1}{(n+2)2^{n+2}}$$



              Then you are done.



              You accidentally added the two middle terms instead of subtracting.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Your error is just after the sixth step from the bottom:



                $$frac{1}2 -frac {1}{(n+1)2^{n+1}} +frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}=frac{1}2 -frac{1}{(n+2)2^{n+2}}$$



                Then you are done.



                You accidentally added the two middle terms instead of subtracting.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Your error is just after the sixth step from the bottom:



                  $$frac{1}2 -frac {1}{(n+1)2^{n+1}} +frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}=frac{1}2 -frac{1}{(n+2)2^{n+2}}$$



                  Then you are done.



                  You accidentally added the two middle terms instead of subtracting.






                  share|cite|improve this answer









                  $endgroup$



                  Your error is just after the sixth step from the bottom:



                  $$frac{1}2 -frac {1}{(n+1)2^{n+1}} +frac{1}{(n+1)2^{n+1}} - frac{1}{(n+2)2^{n+2}}=frac{1}2 -frac{1}{(n+2)2^{n+2}}$$



                  Then you are done.



                  You accidentally added the two middle terms instead of subtracting.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 4 hours ago









                  John Wayland BalesJohn Wayland Bales

                  15.1k21238




                  15.1k21238























                      2












                      $begingroup$

                      Using a telescoping sum, we get
                      $$
                      begin{align}
                      sum_{k=1}^nfrac{k+2}{k(k+1)2^{k+1}}
                      &=sum_{k=1}^nleft(frac1{k2^k}-frac1{(k+1)2^{k+1}}right)\
                      &=sum_{k=1}^nfrac1{k2^k}-sum_{k=2}^{n+1}frac1{k2^k}\
                      &=frac12-frac1{(n+1)2^{n+1}}
                      end{align}
                      $$






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        Using a telescoping sum, we get
                        $$
                        begin{align}
                        sum_{k=1}^nfrac{k+2}{k(k+1)2^{k+1}}
                        &=sum_{k=1}^nleft(frac1{k2^k}-frac1{(k+1)2^{k+1}}right)\
                        &=sum_{k=1}^nfrac1{k2^k}-sum_{k=2}^{n+1}frac1{k2^k}\
                        &=frac12-frac1{(n+1)2^{n+1}}
                        end{align}
                        $$






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          Using a telescoping sum, we get
                          $$
                          begin{align}
                          sum_{k=1}^nfrac{k+2}{k(k+1)2^{k+1}}
                          &=sum_{k=1}^nleft(frac1{k2^k}-frac1{(k+1)2^{k+1}}right)\
                          &=sum_{k=1}^nfrac1{k2^k}-sum_{k=2}^{n+1}frac1{k2^k}\
                          &=frac12-frac1{(n+1)2^{n+1}}
                          end{align}
                          $$






                          share|cite|improve this answer









                          $endgroup$



                          Using a telescoping sum, we get
                          $$
                          begin{align}
                          sum_{k=1}^nfrac{k+2}{k(k+1)2^{k+1}}
                          &=sum_{k=1}^nleft(frac1{k2^k}-frac1{(k+1)2^{k+1}}right)\
                          &=sum_{k=1}^nfrac1{k2^k}-sum_{k=2}^{n+1}frac1{k2^k}\
                          &=frac12-frac1{(n+1)2^{n+1}}
                          end{align}
                          $$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 3 hours ago









                          robjohn♦robjohn

                          270k27312639




                          270k27312639






























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