Proving by induction of $n$ that $sum_{k=1}^n frac {k+2}{k(k+1)2^{k+1}} = frac{1}{2} - frac1{(n+1)2^{n+1}} $

Multi tool use
Multi tool use












5












$begingroup$



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




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











$endgroup$

















    5












    $begingroup$



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




    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











    $endgroup$















      5












      5








      5


      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 18 hours ago









      Asaf Karagila♦

      307k33439770




      307k33439770










      asked 23 hours ago









      BrownieBrownie

      1977




      1977






















          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          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









            $endgroup$













              Your Answer





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              });
              });
              }, "mathjax-editing");

              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "69"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3162553%2fproving-by-induction-of-n-that-sum-k-1n-frac-k2kk12k1-fra%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              4












              $begingroup$

              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









              $endgroup$


















                4












                $begingroup$

                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









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  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 23 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 22 hours ago









                          robjohn♦robjohn

                          270k27312639




                          270k27312639






























                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid …



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3162553%2fproving-by-induction-of-n-that-sum-k-1n-frac-k2kk12k1-fra%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              Zg4Ab,X7GEp hio87iRolJI
                              -
                              1,9,3xKgv wTcShLP eNqvImu N4dZH0fJ,0HMYX SYTANByTbj 7E4DrUIzmrDUA97A84OfWuRaXSHCh EJuetvi

                              Popular posts from this blog

                              Bruad Bilen | Luke uk diar | NawigatsjuunCommonskategorii: BruadCommonskategorii: RunstükenWikiquote: Bruad

                              Image
                              Öömrang ArtiikelKultuurIidj an drankHoonwerk ualhuuchsjiiskualgermaanskurnweederkwegelssees Wikiquote: Bruad (function()var node=document.getElementById("mw-dismissablenotice-anonplace");if(node)node.outerHTML="u003Cdiv class="mw-dismissable-notice"u003Eu003Cdiv class="mw-dismissable-notice-close"u003E[u003Ca tabindex="0" role="button"u003EFersteegu003C/au003E]u003C/divu003Eu003Cdiv class="mw-dismissable-notice-body"u003Eu003Cdiv id="localNotice" lang="frr" dir="ltr"u003Eu003Cdiv id="sitenotice"u003Enu003Ccenteru003Enu003Ctable class="rahmenfarbe4" style="border-style: solid; border-width: thin; width: 70%;"u003Ennu003Ctbodyu003Eu003Ctru003Enu003Ctd style="text-align:center;"u003Eu003Ca href="/wiki/Datei:Nordfriesischeflagge.svg" class="image"u003Eu003Cimg alt="Nordfriesischeflagge.svg" src="//upload.wikimedia.or...
                              Read more

                              Færeyskur hestur Heimild | Tengill | Tilvísanir | LeiðsagnarvalRossið - síða um færeyska hrossið á færeyskuGott ár hjá færeyska hestinum

                              Image
                              FæreyjarHestar færeyskahjaltlandshestinumíslenski hesturinnBretlandskolanámumenskuWikipediafæreyskuWikipediaRossið - síða um færeyska hrossið á færeysku Færeyskur hestur Úr Wikipediu, frjálsa alfræðiritinu Jump to navigation Jump to search Rani og Grani. Grani í vetrarfeldi. Færeyski hesturinn (færeyska: Føroyska rossið ) er hestur sem hefur lifað í Færeyjum í hundruði ára. Hann er smágerður, 120-132 cm á hæð, og helst skyldur hjaltlandshestinum. Hlutverk hans var að draga vagna og plóg og bera klyfjar. Hann hefur fjórar gangtegundir eins og íslenski hesturinn (þar með talið tölt) og er með fjölda litaafbrigði. Nú til dags er hann notaður til frístunda. Í lok 19. aldar voru til um 800 hross. Mö...
                              Read more

                              Chléb Obsah Etymologie | Pojmy při krájení bochníku nebo pecnu chleba | Receptura a druhy | Typy českého chleba | Kvalita chleba v České republice | Cena chleba | Konzumace | Postup výroby | Odkazy | Navigační menuDostupné onlineKdo si mastí kapsu na chlebu? Pekaři to nejsouVývoj spotřebitelských cen – Český statistický úřadDostupné onlineJak se co dělá: Chleba4008364-08669

                              Image
                              Pečivo hovorověpotravinámpečenímpařenímsmaženímtěstamoukyvodyjedlá sůlkvásekkvasnicekořeníkmínsezammáklovci a sběračipitagyrosemGótštinaněmčinaFinštinaestonštinaruštinapolštinagermánských jazykůněmeckynizozemskyšvédskydánskylatinsky12. stoletítěstapšeničnépecilepkupšeničnéžitnéječmennékukuřičnéovesné[1][2]symbolickénáboženstvíchpitamaceskváskupšeničnéžitnéEvropěkváskemFermentacísacharidůcukrukvasinkyoxid uhličitýpekárenkypřícího prášku do pečivasamokynoucí moukypodmáslíjedlou sodukvalityenzymypřídatné látkyžitnénařízení Evropské komise o regulaci karcinogenníh...
                              Read more