Turning product sequences into factorials












1












$begingroup$


I am trying to figure out the steps between these equal expressions in order to get a more general understanding of product sequences:
$$prod_{k=0}^{n}left(3n-kright) + prod_{k=n}^{2n-3}left(2n-kright) = prod_{j=2n}^{3n}j + prod_{j=3}^{n}j =frac{(3n)!}{(2n-1)!}+frac{n!}{2}$$



I know that $ n! :=prod_{k=1}^{n}k$ but I can't figure out how that helps me understand the above equation.



edit: Thank you for the great help! Another thing I don't understand, is how I get from $prod_{k=0}^{n}left(3n-kright) + prod_{k=n}^{2n-3}left(2n-kright)$ to $prod_{j=2n}^{3n}j + prod_{j=3}^{n}j$. Any help with understanding this is much appreciated, I will try to figure it out myself while I wait for answers.






sequences-and-series discrete-mathematics factorial







share|cite|improve this question











$endgroup$












  • $begingroup$
    Its quite straightforward. $displaystyleprod_{j=2n}^{3n}j=(2n)(2n+1)cdots(3n)=dfrac{(2n-1)!:(2n)(2n+1)cdots(3n)}{(2n-1)!}=dfrac{(3n)!}{(2n-1)!} $. Similarly the other summation follows.
    $endgroup$
    – Yadati Kiran
    Dec 19 '18 at 9:30












  • $begingroup$
    Just put the values of $k$ in both the products on the either side and see for yourself! It is nothing. You should be able to do it now! Just keep in mind that $n!=n(n-1)(n-2)...1$
    $endgroup$
    – Sameer Baheti
    Dec 19 '18 at 9:30












  • $begingroup$
    It might also be worth noting that the addition is doing nothing special. There are two terms here, and the equalities hold for the terms separately.
    $endgroup$
    – Mees de Vries
    Dec 19 '18 at 11:05
















1












$begingroup$


I am trying to figure out the steps between these equal expressions in order to get a more general understanding of product sequences:
$$prod_{k=0}^{n}left(3n-kright) + prod_{k=n}^{2n-3}left(2n-kright) = prod_{j=2n}^{3n}j + prod_{j=3}^{n}j =frac{(3n)!}{(2n-1)!}+frac{n!}{2}$$



I know that $ n! :=prod_{k=1}^{n}k$ but I can't figure out how that helps me understand the above equation.



edit: Thank you for the great help! Another thing I don't understand, is how I get from $prod_{k=0}^{n}left(3n-kright) + prod_{k=n}^{2n-3}left(2n-kright)$ to $prod_{j=2n}^{3n}j + prod_{j=3}^{n}j$. Any help with understanding this is much appreciated, I will try to figure it out myself while I wait for answers.






sequences-and-series discrete-mathematics factorial







share|cite|improve this question











$endgroup$












  • $begingroup$
    Its quite straightforward. $displaystyleprod_{j=2n}^{3n}j=(2n)(2n+1)cdots(3n)=dfrac{(2n-1)!:(2n)(2n+1)cdots(3n)}{(2n-1)!}=dfrac{(3n)!}{(2n-1)!} $. Similarly the other summation follows.
    $endgroup$
    – Yadati Kiran
    Dec 19 '18 at 9:30












  • $begingroup$
    Just put the values of $k$ in both the products on the either side and see for yourself! It is nothing. You should be able to do it now! Just keep in mind that $n!=n(n-1)(n-2)...1$
    $endgroup$
    – Sameer Baheti
    Dec 19 '18 at 9:30












  • $begingroup$
    It might also be worth noting that the addition is doing nothing special. There are two terms here, and the equalities hold for the terms separately.
    $endgroup$
    – Mees de Vries
    Dec 19 '18 at 11:05














1












1








1


0



$begingroup$


I am trying to figure out the steps between these equal expressions in order to get a more general understanding of product sequences:
$$prod_{k=0}^{n}left(3n-kright) + prod_{k=n}^{2n-3}left(2n-kright) = prod_{j=2n}^{3n}j + prod_{j=3}^{n}j =frac{(3n)!}{(2n-1)!}+frac{n!}{2}$$



I know that $ n! :=prod_{k=1}^{n}k$ but I can't figure out how that helps me understand the above equation.



edit: Thank you for the great help! Another thing I don't understand, is how I get from $prod_{k=0}^{n}left(3n-kright) + prod_{k=n}^{2n-3}left(2n-kright)$ to $prod_{j=2n}^{3n}j + prod_{j=3}^{n}j$. Any help with understanding this is much appreciated, I will try to figure it out myself while I wait for answers.






sequences-and-series discrete-mathematics factorial







share|cite|improve this question











$endgroup$




I am trying to figure out the steps between these equal expressions in order to get a more general understanding of product sequences:
$$prod_{k=0}^{n}left(3n-kright) + prod_{k=n}^{2n-3}left(2n-kright) = prod_{j=2n}^{3n}j + prod_{j=3}^{n}j =frac{(3n)!}{(2n-1)!}+frac{n!}{2}$$



I know that $ n! :=prod_{k=1}^{n}k$ but I can't figure out how that helps me understand the above equation.



edit: Thank you for the great help! Another thing I don't understand, is how I get from $prod_{k=0}^{n}left(3n-kright) + prod_{k=n}^{2n-3}left(2n-kright)$ to $prod_{j=2n}^{3n}j + prod_{j=3}^{n}j$. Any help with understanding this is much appreciated, I will try to figure it out myself while I wait for answers.






sequences-and-series discrete-mathematics factorial




sequences-and-series discrete-mathematics factorial






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 10:47







Lucky

















asked Dec 19 '18 at 9:22









LuckyLucky

1177




1177












  • $begingroup$
    Its quite straightforward. $displaystyleprod_{j=2n}^{3n}j=(2n)(2n+1)cdots(3n)=dfrac{(2n-1)!:(2n)(2n+1)cdots(3n)}{(2n-1)!}=dfrac{(3n)!}{(2n-1)!} $. Similarly the other summation follows.
    $endgroup$
    – Yadati Kiran
    Dec 19 '18 at 9:30












  • $begingroup$
    Just put the values of $k$ in both the products on the either side and see for yourself! It is nothing. You should be able to do it now! Just keep in mind that $n!=n(n-1)(n-2)...1$
    $endgroup$
    – Sameer Baheti
    Dec 19 '18 at 9:30












  • $begingroup$
    It might also be worth noting that the addition is doing nothing special. There are two terms here, and the equalities hold for the terms separately.
    $endgroup$
    – Mees de Vries
    Dec 19 '18 at 11:05


















  • $begingroup$
    Its quite straightforward. $displaystyleprod_{j=2n}^{3n}j=(2n)(2n+1)cdots(3n)=dfrac{(2n-1)!:(2n)(2n+1)cdots(3n)}{(2n-1)!}=dfrac{(3n)!}{(2n-1)!} $. Similarly the other summation follows.
    $endgroup$
    – Yadati Kiran
    Dec 19 '18 at 9:30












  • $begingroup$
    Just put the values of $k$ in both the products on the either side and see for yourself! It is nothing. You should be able to do it now! Just keep in mind that $n!=n(n-1)(n-2)...1$
    $endgroup$
    – Sameer Baheti
    Dec 19 '18 at 9:30












  • $begingroup$
    It might also be worth noting that the addition is doing nothing special. There are two terms here, and the equalities hold for the terms separately.
    $endgroup$
    – Mees de Vries
    Dec 19 '18 at 11:05
















$begingroup$
Its quite straightforward. $displaystyleprod_{j=2n}^{3n}j=(2n)(2n+1)cdots(3n)=dfrac{(2n-1)!:(2n)(2n+1)cdots(3n)}{(2n-1)!}=dfrac{(3n)!}{(2n-1)!} $. Similarly the other summation follows.
$endgroup$
– Yadati Kiran
Dec 19 '18 at 9:30






$begingroup$
Its quite straightforward. $displaystyleprod_{j=2n}^{3n}j=(2n)(2n+1)cdots(3n)=dfrac{(2n-1)!:(2n)(2n+1)cdots(3n)}{(2n-1)!}=dfrac{(3n)!}{(2n-1)!} $. Similarly the other summation follows.
$endgroup$
– Yadati Kiran
Dec 19 '18 at 9:30














$begingroup$
Just put the values of $k$ in both the products on the either side and see for yourself! It is nothing. You should be able to do it now! Just keep in mind that $n!=n(n-1)(n-2)...1$
$endgroup$
– Sameer Baheti
Dec 19 '18 at 9:30






$begingroup$
Just put the values of $k$ in both the products on the either side and see for yourself! It is nothing. You should be able to do it now! Just keep in mind that $n!=n(n-1)(n-2)...1$
$endgroup$
– Sameer Baheti
Dec 19 '18 at 9:30














$begingroup$
It might also be worth noting that the addition is doing nothing special. There are two terms here, and the equalities hold for the terms separately.
$endgroup$
– Mees de Vries
Dec 19 '18 at 11:05




$begingroup$
It might also be worth noting that the addition is doing nothing special. There are two terms here, and the equalities hold for the terms separately.
$endgroup$
– Mees de Vries
Dec 19 '18 at 11:05










2 Answers
2






active

oldest

votes


















1












$begingroup$

Let's evaluate a few terms for $$prod_{k=0}^n (3n-k)$$



when $k=0$, we are multipliying $3n$.



when $k=1$, we are multipliying $3n-1$.



when $k=2$, we are multipliying $3n-2$.



and so on. Each time, the individual term reduces by $1$.



when $k=n$, we are multiplying $3n-n=2n$



Hence we are just multiplying every term from $2n$ to $3n$.



$$prod_{k=0}^n (3n-k)=prod_{j=2n}^{3n}j=frac{(3n)!}{(2n-1)!}$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$


    We obtain for integers $ngeq 2$:



    begin{align*}
    color{blue}{prod_{k=0}^{n}}&color{blue}{left(3n-kright) + prod_{k=n}^{2n-3}left(2n-kright)}\
    &=prod_{k=0}^n(2n+k)+prod_{k=n}^{2n-3}(k+3-n)tag{1}\
    &,,color{blue}{=prod_{k=2n}^{3n}k+prod_{k=3}^{n}k}tag{2}\
    &=left(prod_{k=1}^{3n}kright)/left(prod_{k=1}^{2n-1}kright)+left(prod_{k=1}^n kright)/left(prod_{k=1}^2kright)tag{3}\
    &,,color{blue}{=frac{(3n)!}{(2n-1)!}+frac{n!}{2}}
    end{align*}




    Comment:




    • In (1) we change in both products the order of multiplication: $kto (n-k)$ and $kto ((2n-3)-k+n)$.


    • In (2) we shift the left product by $2n$ via $kto2n+k$ and we shift the right product by $-n+3$ via $kto k+n-3$.


    • In (3) we complete the product to start with factors from $1$ and compensate this with the product in the denominator.



    We start with $ngeq 2$, since the upper limit of the right product in (1) is negative in case $n=1$ invalidating the claim.






    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%2f3046205%2fturning-product-sequences-into-factorials%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









      1












      $begingroup$

      Let's evaluate a few terms for $$prod_{k=0}^n (3n-k)$$



      when $k=0$, we are multipliying $3n$.



      when $k=1$, we are multipliying $3n-1$.



      when $k=2$, we are multipliying $3n-2$.



      and so on. Each time, the individual term reduces by $1$.



      when $k=n$, we are multiplying $3n-n=2n$



      Hence we are just multiplying every term from $2n$ to $3n$.



      $$prod_{k=0}^n (3n-k)=prod_{j=2n}^{3n}j=frac{(3n)!}{(2n-1)!}$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Let's evaluate a few terms for $$prod_{k=0}^n (3n-k)$$



        when $k=0$, we are multipliying $3n$.



        when $k=1$, we are multipliying $3n-1$.



        when $k=2$, we are multipliying $3n-2$.



        and so on. Each time, the individual term reduces by $1$.



        when $k=n$, we are multiplying $3n-n=2n$



        Hence we are just multiplying every term from $2n$ to $3n$.



        $$prod_{k=0}^n (3n-k)=prod_{j=2n}^{3n}j=frac{(3n)!}{(2n-1)!}$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Let's evaluate a few terms for $$prod_{k=0}^n (3n-k)$$



          when $k=0$, we are multipliying $3n$.



          when $k=1$, we are multipliying $3n-1$.



          when $k=2$, we are multipliying $3n-2$.



          and so on. Each time, the individual term reduces by $1$.



          when $k=n$, we are multiplying $3n-n=2n$



          Hence we are just multiplying every term from $2n$ to $3n$.



          $$prod_{k=0}^n (3n-k)=prod_{j=2n}^{3n}j=frac{(3n)!}{(2n-1)!}$$






          share|cite|improve this answer









          $endgroup$



          Let's evaluate a few terms for $$prod_{k=0}^n (3n-k)$$



          when $k=0$, we are multipliying $3n$.



          when $k=1$, we are multipliying $3n-1$.



          when $k=2$, we are multipliying $3n-2$.



          and so on. Each time, the individual term reduces by $1$.



          when $k=n$, we are multiplying $3n-n=2n$



          Hence we are just multiplying every term from $2n$ to $3n$.



          $$prod_{k=0}^n (3n-k)=prod_{j=2n}^{3n}j=frac{(3n)!}{(2n-1)!}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 19 '18 at 12:02









          Siong Thye GohSiong Thye Goh

          102k1466118




          102k1466118























              1












              $begingroup$


              We obtain for integers $ngeq 2$:



              begin{align*}
              color{blue}{prod_{k=0}^{n}}&color{blue}{left(3n-kright) + prod_{k=n}^{2n-3}left(2n-kright)}\
              &=prod_{k=0}^n(2n+k)+prod_{k=n}^{2n-3}(k+3-n)tag{1}\
              &,,color{blue}{=prod_{k=2n}^{3n}k+prod_{k=3}^{n}k}tag{2}\
              &=left(prod_{k=1}^{3n}kright)/left(prod_{k=1}^{2n-1}kright)+left(prod_{k=1}^n kright)/left(prod_{k=1}^2kright)tag{3}\
              &,,color{blue}{=frac{(3n)!}{(2n-1)!}+frac{n!}{2}}
              end{align*}




              Comment:




              • In (1) we change in both products the order of multiplication: $kto (n-k)$ and $kto ((2n-3)-k+n)$.


              • In (2) we shift the left product by $2n$ via $kto2n+k$ and we shift the right product by $-n+3$ via $kto k+n-3$.


              • In (3) we complete the product to start with factors from $1$ and compensate this with the product in the denominator.



              We start with $ngeq 2$, since the upper limit of the right product in (1) is negative in case $n=1$ invalidating the claim.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$


                We obtain for integers $ngeq 2$:



                begin{align*}
                color{blue}{prod_{k=0}^{n}}&color{blue}{left(3n-kright) + prod_{k=n}^{2n-3}left(2n-kright)}\
                &=prod_{k=0}^n(2n+k)+prod_{k=n}^{2n-3}(k+3-n)tag{1}\
                &,,color{blue}{=prod_{k=2n}^{3n}k+prod_{k=3}^{n}k}tag{2}\
                &=left(prod_{k=1}^{3n}kright)/left(prod_{k=1}^{2n-1}kright)+left(prod_{k=1}^n kright)/left(prod_{k=1}^2kright)tag{3}\
                &,,color{blue}{=frac{(3n)!}{(2n-1)!}+frac{n!}{2}}
                end{align*}




                Comment:




                • In (1) we change in both products the order of multiplication: $kto (n-k)$ and $kto ((2n-3)-k+n)$.


                • In (2) we shift the left product by $2n$ via $kto2n+k$ and we shift the right product by $-n+3$ via $kto k+n-3$.


                • In (3) we complete the product to start with factors from $1$ and compensate this with the product in the denominator.



                We start with $ngeq 2$, since the upper limit of the right product in (1) is negative in case $n=1$ invalidating the claim.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$


                  We obtain for integers $ngeq 2$:



                  begin{align*}
                  color{blue}{prod_{k=0}^{n}}&color{blue}{left(3n-kright) + prod_{k=n}^{2n-3}left(2n-kright)}\
                  &=prod_{k=0}^n(2n+k)+prod_{k=n}^{2n-3}(k+3-n)tag{1}\
                  &,,color{blue}{=prod_{k=2n}^{3n}k+prod_{k=3}^{n}k}tag{2}\
                  &=left(prod_{k=1}^{3n}kright)/left(prod_{k=1}^{2n-1}kright)+left(prod_{k=1}^n kright)/left(prod_{k=1}^2kright)tag{3}\
                  &,,color{blue}{=frac{(3n)!}{(2n-1)!}+frac{n!}{2}}
                  end{align*}




                  Comment:




                  • In (1) we change in both products the order of multiplication: $kto (n-k)$ and $kto ((2n-3)-k+n)$.


                  • In (2) we shift the left product by $2n$ via $kto2n+k$ and we shift the right product by $-n+3$ via $kto k+n-3$.


                  • In (3) we complete the product to start with factors from $1$ and compensate this with the product in the denominator.



                  We start with $ngeq 2$, since the upper limit of the right product in (1) is negative in case $n=1$ invalidating the claim.






                  share|cite|improve this answer









                  $endgroup$




                  We obtain for integers $ngeq 2$:



                  begin{align*}
                  color{blue}{prod_{k=0}^{n}}&color{blue}{left(3n-kright) + prod_{k=n}^{2n-3}left(2n-kright)}\
                  &=prod_{k=0}^n(2n+k)+prod_{k=n}^{2n-3}(k+3-n)tag{1}\
                  &,,color{blue}{=prod_{k=2n}^{3n}k+prod_{k=3}^{n}k}tag{2}\
                  &=left(prod_{k=1}^{3n}kright)/left(prod_{k=1}^{2n-1}kright)+left(prod_{k=1}^n kright)/left(prod_{k=1}^2kright)tag{3}\
                  &,,color{blue}{=frac{(3n)!}{(2n-1)!}+frac{n!}{2}}
                  end{align*}




                  Comment:




                  • In (1) we change in both products the order of multiplication: $kto (n-k)$ and $kto ((2n-3)-k+n)$.


                  • In (2) we shift the left product by $2n$ via $kto2n+k$ and we shift the right product by $-n+3$ via $kto k+n-3$.


                  • In (3) we complete the product to start with factors from $1$ and compensate this with the product in the denominator.



                  We start with $ngeq 2$, since the upper limit of the right product in (1) is negative in case $n=1$ invalidating the claim.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 19 '18 at 22:20









                  Markus ScheuerMarkus Scheuer

                  62.2k459149




                  62.2k459149






























                      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%2f3046205%2fturning-product-sequences-into-factorials%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

                      Index of /

                      This page is only for reference, If you need detailed information, please check here
                      Read more

                      Tribalistas

                      Image
                        Nota: Para o álbum homônimo, veja Tribalistas (álbum de 2002). Para o segundo álbum homônimo, veja Tribalistas (álbum de 2017). Para o tipo de organização tribal, veja Tribalismo. Tribalistas Informação geral Origem Salvador e Rio de Janeiro País   Brasil Gênero(s) MPB Pop Samba Período em atividade 2002–2004 2017–atualmente Gravadora(s) EMI Phonomotor Records Universal Music Afiliação(ões) Dadi Pretinho da Serrinha Margareth Menezes Integrantes Arnaldo Antunes Carlinhos Brown Marisa Monte Página oficial www2.uol.com.br/tribalistas/ ...
                      Read more

                      Listed building

                      Image
                      Listed building (literalmente: prédio listado ) é toda estrutura protegida por lei pelo complexo sistema de tombamento do Reino Unido. Um listed building é qualquer estrutura oficialmente designada como tendo especial interesse arquitetônico, histórico ou cultural. É um estatuto amplamente usado, aplicado a centenas de milhares de estruturas. Estruturas tombadas não podem ser alteradas sem permissão especial, e desfrutam de isenções fiscais e outros benefícios. Cada um dos países do Reino Unido trata tombamento de forma diferente: Índice 1 Inglaterra 2 País de Gales 3 Irlanda do Norte 4 Escócia 5 Referências Inglaterra | O Palácio de Buckingham, Londres. As estruturas tombadas são listadas no Statutory List of Buildings of Special Architectural or Historic Interest , e incluem campos, sinais de trânsito, piers e áreas geográficas. A principal legislação pertinente é o Ancient Monuments and Archaeological Areas Act 1979 , [ 1 ] embora legisla...
                      Read more