Prove $sum_{r=2}^n {n choose r} r(r-1) = n(n-1)2^{n-2}$ for $ngeq 2$












1












$begingroup$


How to prove following:



$$sum_{r=2}^n {n choose r} r(r-1) = n(n-1)2^{n-2}$$ for $ngeq 2$



Thanks!!






discrete-mathematics proof-explanation







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I'd use induction. Start with $n = 2$ and then prove that if the statement is valid for $n$ it is also valid for $n+1$. Basically, people get more enthusiastic about questions if the poster shows hir/her efforts
    $endgroup$
    – Ronald
    Dec 6 '18 at 10:17
















1












$begingroup$


How to prove following:



$$sum_{r=2}^n {n choose r} r(r-1) = n(n-1)2^{n-2}$$ for $ngeq 2$



Thanks!!






discrete-mathematics proof-explanation







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I'd use induction. Start with $n = 2$ and then prove that if the statement is valid for $n$ it is also valid for $n+1$. Basically, people get more enthusiastic about questions if the poster shows hir/her efforts
    $endgroup$
    – Ronald
    Dec 6 '18 at 10:17














1












1








1





$begingroup$


How to prove following:



$$sum_{r=2}^n {n choose r} r(r-1) = n(n-1)2^{n-2}$$ for $ngeq 2$



Thanks!!






discrete-mathematics proof-explanation







share|cite|improve this question











$endgroup$




How to prove following:



$$sum_{r=2}^n {n choose r} r(r-1) = n(n-1)2^{n-2}$$ for $ngeq 2$



Thanks!!






discrete-mathematics proof-explanation




discrete-mathematics proof-explanation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 10:21









greedoid

38.7k114797




38.7k114797










asked Dec 6 '18 at 10:15









TimgascdTimgascd

303




303








  • 1




    $begingroup$
    I'd use induction. Start with $n = 2$ and then prove that if the statement is valid for $n$ it is also valid for $n+1$. Basically, people get more enthusiastic about questions if the poster shows hir/her efforts
    $endgroup$
    – Ronald
    Dec 6 '18 at 10:17














  • 1




    $begingroup$
    I'd use induction. Start with $n = 2$ and then prove that if the statement is valid for $n$ it is also valid for $n+1$. Basically, people get more enthusiastic about questions if the poster shows hir/her efforts
    $endgroup$
    – Ronald
    Dec 6 '18 at 10:17








1




1




$begingroup$
I'd use induction. Start with $n = 2$ and then prove that if the statement is valid for $n$ it is also valid for $n+1$. Basically, people get more enthusiastic about questions if the poster shows hir/her efforts
$endgroup$
– Ronald
Dec 6 '18 at 10:17




$begingroup$
I'd use induction. Start with $n = 2$ and then prove that if the statement is valid for $n$ it is also valid for $n+1$. Basically, people get more enthusiastic about questions if the poster shows hir/her efforts
$endgroup$
– Ronald
Dec 6 '18 at 10:17










3 Answers
3






active

oldest

votes


















0












$begingroup$

Hint:



For $r(r-1)ne0$



$$r(r-1)binom nr=r(r-1)n(n-1)dfrac{(n-2)!}{r(r-1)cdot(r-2)!cdot{n-2-(r-2)}!}=n(n-1)binom{n-2}{r-2}$$



Now in
$$(a+b)^m=sum_{r=0}^mbinom mr a^{m-r}b^r$$



put $a=b=1, m=n-2$



Some observations :




  1. $$sum_{r=2}^nbinom nrr(r-1)=sum_{r=0}^nbinom nrr(r-1)$$


  2. The proposition trivially holds true for $n=0,1$







share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry, I do not really understand all of them. Can you explain it in detail? Thanks
    $endgroup$
    – Timgascd
    Dec 6 '18 at 10:41










  • $begingroup$
    @Hgascd, All of them or none of them? $$binom nr=dfrac{n!}{(n-r)! r!}=dfrac{n(n-1)}{r(r-1)}cdotdfrac{(n-2)!}{(r-2)!cdot{n-2-(r-2)}!}=$$
    $endgroup$
    – lab bhattacharjee
    Dec 6 '18 at 10:55












  • $begingroup$
    Thanks a lot. I got it
    $endgroup$
    – Timgascd
    Dec 6 '18 at 11:04



















1












$begingroup$

Question: On how many ways can we choose a group in a set of $n$ people and then president and then vicepresident?



Well we can first a group of $r$ people, that is ${nchoose r}$, for every $rleq n$, and then president among them, so we have $r$ choises and then $r-1$ choises for V.P. Suming (we can sum from $0$) for all $r$ we get:



$$sum_{r=0}^{n} r(r-1)C_{n}^{r} $$



On the other hand we can first choose a president among all people, so we have $n$ posibilities and then V.P. for who we have $n-1$ choises and then we choose any set in set of $n-2$ people, for that we have $2^{n-2}$ choises, so:



$$n(n-1)2^{n-2}$$ and this is the answer to your question.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Hint: Take $f(x)=(1+x)^n$ and consider $f''(1)$.






    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%2f3028304%2fprove-sum-r-2n-n-choose-r-rr-1-nn-12n-2-for-n-geq-2%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      Hint:



      For $r(r-1)ne0$



      $$r(r-1)binom nr=r(r-1)n(n-1)dfrac{(n-2)!}{r(r-1)cdot(r-2)!cdot{n-2-(r-2)}!}=n(n-1)binom{n-2}{r-2}$$



      Now in
      $$(a+b)^m=sum_{r=0}^mbinom mr a^{m-r}b^r$$



      put $a=b=1, m=n-2$



      Some observations :




      1. $$sum_{r=2}^nbinom nrr(r-1)=sum_{r=0}^nbinom nrr(r-1)$$


      2. The proposition trivially holds true for $n=0,1$







      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Sorry, I do not really understand all of them. Can you explain it in detail? Thanks
        $endgroup$
        – Timgascd
        Dec 6 '18 at 10:41










      • $begingroup$
        @Hgascd, All of them or none of them? $$binom nr=dfrac{n!}{(n-r)! r!}=dfrac{n(n-1)}{r(r-1)}cdotdfrac{(n-2)!}{(r-2)!cdot{n-2-(r-2)}!}=$$
        $endgroup$
        – lab bhattacharjee
        Dec 6 '18 at 10:55












      • $begingroup$
        Thanks a lot. I got it
        $endgroup$
        – Timgascd
        Dec 6 '18 at 11:04
















      0












      $begingroup$

      Hint:



      For $r(r-1)ne0$



      $$r(r-1)binom nr=r(r-1)n(n-1)dfrac{(n-2)!}{r(r-1)cdot(r-2)!cdot{n-2-(r-2)}!}=n(n-1)binom{n-2}{r-2}$$



      Now in
      $$(a+b)^m=sum_{r=0}^mbinom mr a^{m-r}b^r$$



      put $a=b=1, m=n-2$



      Some observations :




      1. $$sum_{r=2}^nbinom nrr(r-1)=sum_{r=0}^nbinom nrr(r-1)$$


      2. The proposition trivially holds true for $n=0,1$







      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Sorry, I do not really understand all of them. Can you explain it in detail? Thanks
        $endgroup$
        – Timgascd
        Dec 6 '18 at 10:41










      • $begingroup$
        @Hgascd, All of them or none of them? $$binom nr=dfrac{n!}{(n-r)! r!}=dfrac{n(n-1)}{r(r-1)}cdotdfrac{(n-2)!}{(r-2)!cdot{n-2-(r-2)}!}=$$
        $endgroup$
        – lab bhattacharjee
        Dec 6 '18 at 10:55












      • $begingroup$
        Thanks a lot. I got it
        $endgroup$
        – Timgascd
        Dec 6 '18 at 11:04














      0












      0








      0





      $begingroup$

      Hint:



      For $r(r-1)ne0$



      $$r(r-1)binom nr=r(r-1)n(n-1)dfrac{(n-2)!}{r(r-1)cdot(r-2)!cdot{n-2-(r-2)}!}=n(n-1)binom{n-2}{r-2}$$



      Now in
      $$(a+b)^m=sum_{r=0}^mbinom mr a^{m-r}b^r$$



      put $a=b=1, m=n-2$



      Some observations :




      1. $$sum_{r=2}^nbinom nrr(r-1)=sum_{r=0}^nbinom nrr(r-1)$$


      2. The proposition trivially holds true for $n=0,1$







      share|cite|improve this answer









      $endgroup$



      Hint:



      For $r(r-1)ne0$



      $$r(r-1)binom nr=r(r-1)n(n-1)dfrac{(n-2)!}{r(r-1)cdot(r-2)!cdot{n-2-(r-2)}!}=n(n-1)binom{n-2}{r-2}$$



      Now in
      $$(a+b)^m=sum_{r=0}^mbinom mr a^{m-r}b^r$$



      put $a=b=1, m=n-2$



      Some observations :




      1. $$sum_{r=2}^nbinom nrr(r-1)=sum_{r=0}^nbinom nrr(r-1)$$


      2. The proposition trivially holds true for $n=0,1$








      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 6 '18 at 10:20









      lab bhattacharjeelab bhattacharjee

      224k15156274




      224k15156274












      • $begingroup$
        Sorry, I do not really understand all of them. Can you explain it in detail? Thanks
        $endgroup$
        – Timgascd
        Dec 6 '18 at 10:41










      • $begingroup$
        @Hgascd, All of them or none of them? $$binom nr=dfrac{n!}{(n-r)! r!}=dfrac{n(n-1)}{r(r-1)}cdotdfrac{(n-2)!}{(r-2)!cdot{n-2-(r-2)}!}=$$
        $endgroup$
        – lab bhattacharjee
        Dec 6 '18 at 10:55












      • $begingroup$
        Thanks a lot. I got it
        $endgroup$
        – Timgascd
        Dec 6 '18 at 11:04


















      • $begingroup$
        Sorry, I do not really understand all of them. Can you explain it in detail? Thanks
        $endgroup$
        – Timgascd
        Dec 6 '18 at 10:41










      • $begingroup$
        @Hgascd, All of them or none of them? $$binom nr=dfrac{n!}{(n-r)! r!}=dfrac{n(n-1)}{r(r-1)}cdotdfrac{(n-2)!}{(r-2)!cdot{n-2-(r-2)}!}=$$
        $endgroup$
        – lab bhattacharjee
        Dec 6 '18 at 10:55












      • $begingroup$
        Thanks a lot. I got it
        $endgroup$
        – Timgascd
        Dec 6 '18 at 11:04
















      $begingroup$
      Sorry, I do not really understand all of them. Can you explain it in detail? Thanks
      $endgroup$
      – Timgascd
      Dec 6 '18 at 10:41




      $begingroup$
      Sorry, I do not really understand all of them. Can you explain it in detail? Thanks
      $endgroup$
      – Timgascd
      Dec 6 '18 at 10:41












      $begingroup$
      @Hgascd, All of them or none of them? $$binom nr=dfrac{n!}{(n-r)! r!}=dfrac{n(n-1)}{r(r-1)}cdotdfrac{(n-2)!}{(r-2)!cdot{n-2-(r-2)}!}=$$
      $endgroup$
      – lab bhattacharjee
      Dec 6 '18 at 10:55






      $begingroup$
      @Hgascd, All of them or none of them? $$binom nr=dfrac{n!}{(n-r)! r!}=dfrac{n(n-1)}{r(r-1)}cdotdfrac{(n-2)!}{(r-2)!cdot{n-2-(r-2)}!}=$$
      $endgroup$
      – lab bhattacharjee
      Dec 6 '18 at 10:55














      $begingroup$
      Thanks a lot. I got it
      $endgroup$
      – Timgascd
      Dec 6 '18 at 11:04




      $begingroup$
      Thanks a lot. I got it
      $endgroup$
      – Timgascd
      Dec 6 '18 at 11:04











      1












      $begingroup$

      Question: On how many ways can we choose a group in a set of $n$ people and then president and then vicepresident?



      Well we can first a group of $r$ people, that is ${nchoose r}$, for every $rleq n$, and then president among them, so we have $r$ choises and then $r-1$ choises for V.P. Suming (we can sum from $0$) for all $r$ we get:



      $$sum_{r=0}^{n} r(r-1)C_{n}^{r} $$



      On the other hand we can first choose a president among all people, so we have $n$ posibilities and then V.P. for who we have $n-1$ choises and then we choose any set in set of $n-2$ people, for that we have $2^{n-2}$ choises, so:



      $$n(n-1)2^{n-2}$$ and this is the answer to your question.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Question: On how many ways can we choose a group in a set of $n$ people and then president and then vicepresident?



        Well we can first a group of $r$ people, that is ${nchoose r}$, for every $rleq n$, and then president among them, so we have $r$ choises and then $r-1$ choises for V.P. Suming (we can sum from $0$) for all $r$ we get:



        $$sum_{r=0}^{n} r(r-1)C_{n}^{r} $$



        On the other hand we can first choose a president among all people, so we have $n$ posibilities and then V.P. for who we have $n-1$ choises and then we choose any set in set of $n-2$ people, for that we have $2^{n-2}$ choises, so:



        $$n(n-1)2^{n-2}$$ and this is the answer to your question.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Question: On how many ways can we choose a group in a set of $n$ people and then president and then vicepresident?



          Well we can first a group of $r$ people, that is ${nchoose r}$, for every $rleq n$, and then president among them, so we have $r$ choises and then $r-1$ choises for V.P. Suming (we can sum from $0$) for all $r$ we get:



          $$sum_{r=0}^{n} r(r-1)C_{n}^{r} $$



          On the other hand we can first choose a president among all people, so we have $n$ posibilities and then V.P. for who we have $n-1$ choises and then we choose any set in set of $n-2$ people, for that we have $2^{n-2}$ choises, so:



          $$n(n-1)2^{n-2}$$ and this is the answer to your question.






          share|cite|improve this answer









          $endgroup$



          Question: On how many ways can we choose a group in a set of $n$ people and then president and then vicepresident?



          Well we can first a group of $r$ people, that is ${nchoose r}$, for every $rleq n$, and then president among them, so we have $r$ choises and then $r-1$ choises for V.P. Suming (we can sum from $0$) for all $r$ we get:



          $$sum_{r=0}^{n} r(r-1)C_{n}^{r} $$



          On the other hand we can first choose a president among all people, so we have $n$ posibilities and then V.P. for who we have $n-1$ choises and then we choose any set in set of $n-2$ people, for that we have $2^{n-2}$ choises, so:



          $$n(n-1)2^{n-2}$$ and this is the answer to your question.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 6 '18 at 10:20









          greedoidgreedoid

          38.7k114797




          38.7k114797























              0












              $begingroup$

              Hint: Take $f(x)=(1+x)^n$ and consider $f''(1)$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Hint: Take $f(x)=(1+x)^n$ and consider $f''(1)$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Hint: Take $f(x)=(1+x)^n$ and consider $f''(1)$.






                  share|cite|improve this answer









                  $endgroup$



                  Hint: Take $f(x)=(1+x)^n$ and consider $f''(1)$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 6 '18 at 10:20









                  lhflhf

                  163k10168388




                  163k10168388






























                      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%2f3028304%2fprove-sum-r-2n-n-choose-r-rr-1-nn-12n-2-for-n-geq-2%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







                      -

                      Popular posts from this blog

                      Berounka

                      Image
                      Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
                      Read more

                      Sphinx de Gizeh

                      Image
                      Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
                      Read more

                      Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

                      Image
                      3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
                      Read more