How to prove $binom{n}{4}=sum_{i=3}^{n-1} binom{i}{3}$?












0












$begingroup$


I was trying to find the number of diagonals of a nth polygon by deduction. Let $a_n$ be the number of the diagonals of thr nth polygon. Then I've got $$a_{n+1}=a_n+binom{n}{3}$$
$$a_n=sum_{i=3}^{n-1} binom{i}{3}$$
But we simply know that $a_n=binom{n}{4}$, where $a_n$ is the number of diagonals.
So we get $binom{n}{4}=sum_{i=3}^{n-2} binom{i}{3}$. My question is whether it is a trivial equation or it can be generalized like $binom{n}{k}=sum_{i=k-1}^{n-1} binom{i}{k-1}$. If it can be, how to prove that?






combinatorics







share|cite|improve this question









$endgroup$












  • $begingroup$
    $$binom nk-binom{n-1}k=?$$ math.stackexchange.com/questions/1125923/…
    $endgroup$
    – lab bhattacharjee
    Jan 15 at 14:03








  • 4




    $begingroup$
    This is the Hockey-stick_identity: see en.wikipedia.org/wiki/Hockey-stick_identity
    $endgroup$
    – Robert Z
    Jan 15 at 14:03










  • $begingroup$
    Can be proved by induction.
    $endgroup$
    – gandalf61
    Jan 15 at 14:11
















0












$begingroup$


I was trying to find the number of diagonals of a nth polygon by deduction. Let $a_n$ be the number of the diagonals of thr nth polygon. Then I've got $$a_{n+1}=a_n+binom{n}{3}$$
$$a_n=sum_{i=3}^{n-1} binom{i}{3}$$
But we simply know that $a_n=binom{n}{4}$, where $a_n$ is the number of diagonals.
So we get $binom{n}{4}=sum_{i=3}^{n-2} binom{i}{3}$. My question is whether it is a trivial equation or it can be generalized like $binom{n}{k}=sum_{i=k-1}^{n-1} binom{i}{k-1}$. If it can be, how to prove that?






combinatorics







share|cite|improve this question









$endgroup$












  • $begingroup$
    $$binom nk-binom{n-1}k=?$$ math.stackexchange.com/questions/1125923/…
    $endgroup$
    – lab bhattacharjee
    Jan 15 at 14:03








  • 4




    $begingroup$
    This is the Hockey-stick_identity: see en.wikipedia.org/wiki/Hockey-stick_identity
    $endgroup$
    – Robert Z
    Jan 15 at 14:03










  • $begingroup$
    Can be proved by induction.
    $endgroup$
    – gandalf61
    Jan 15 at 14:11














0












0








0





$begingroup$


I was trying to find the number of diagonals of a nth polygon by deduction. Let $a_n$ be the number of the diagonals of thr nth polygon. Then I've got $$a_{n+1}=a_n+binom{n}{3}$$
$$a_n=sum_{i=3}^{n-1} binom{i}{3}$$
But we simply know that $a_n=binom{n}{4}$, where $a_n$ is the number of diagonals.
So we get $binom{n}{4}=sum_{i=3}^{n-2} binom{i}{3}$. My question is whether it is a trivial equation or it can be generalized like $binom{n}{k}=sum_{i=k-1}^{n-1} binom{i}{k-1}$. If it can be, how to prove that?






combinatorics







share|cite|improve this question









$endgroup$




I was trying to find the number of diagonals of a nth polygon by deduction. Let $a_n$ be the number of the diagonals of thr nth polygon. Then I've got $$a_{n+1}=a_n+binom{n}{3}$$
$$a_n=sum_{i=3}^{n-1} binom{i}{3}$$
But we simply know that $a_n=binom{n}{4}$, where $a_n$ is the number of diagonals.
So we get $binom{n}{4}=sum_{i=3}^{n-2} binom{i}{3}$. My question is whether it is a trivial equation or it can be generalized like $binom{n}{k}=sum_{i=k-1}^{n-1} binom{i}{k-1}$. If it can be, how to prove that?






combinatorics




combinatorics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 15 at 14:00









Mahmud Nabil SnigdhoMahmud Nabil Snigdho

12




12












  • $begingroup$
    $$binom nk-binom{n-1}k=?$$ math.stackexchange.com/questions/1125923/…
    $endgroup$
    – lab bhattacharjee
    Jan 15 at 14:03








  • 4




    $begingroup$
    This is the Hockey-stick_identity: see en.wikipedia.org/wiki/Hockey-stick_identity
    $endgroup$
    – Robert Z
    Jan 15 at 14:03










  • $begingroup$
    Can be proved by induction.
    $endgroup$
    – gandalf61
    Jan 15 at 14:11


















  • $begingroup$
    $$binom nk-binom{n-1}k=?$$ math.stackexchange.com/questions/1125923/…
    $endgroup$
    – lab bhattacharjee
    Jan 15 at 14:03








  • 4




    $begingroup$
    This is the Hockey-stick_identity: see en.wikipedia.org/wiki/Hockey-stick_identity
    $endgroup$
    – Robert Z
    Jan 15 at 14:03










  • $begingroup$
    Can be proved by induction.
    $endgroup$
    – gandalf61
    Jan 15 at 14:11
















$begingroup$
$$binom nk-binom{n-1}k=?$$ math.stackexchange.com/questions/1125923/…
$endgroup$
– lab bhattacharjee
Jan 15 at 14:03






$begingroup$
$$binom nk-binom{n-1}k=?$$ math.stackexchange.com/questions/1125923/…
$endgroup$
– lab bhattacharjee
Jan 15 at 14:03






4




4




$begingroup$
This is the Hockey-stick_identity: see en.wikipedia.org/wiki/Hockey-stick_identity
$endgroup$
– Robert Z
Jan 15 at 14:03




$begingroup$
This is the Hockey-stick_identity: see en.wikipedia.org/wiki/Hockey-stick_identity
$endgroup$
– Robert Z
Jan 15 at 14:03












$begingroup$
Can be proved by induction.
$endgroup$
– gandalf61
Jan 15 at 14:11




$begingroup$
Can be proved by induction.
$endgroup$
– gandalf61
Jan 15 at 14:11










3 Answers
3






active

oldest

votes


















3












$begingroup$

You may transform the sum into a telescoping sum using the "binomial" fact




  • $binom{i}{3} = binom{i+1}{4} - binom{i}{4}$


Hence,
begin{eqnarray*} sum_{i=3}^{n-1} binom{i}{3}
& = & sum_{i=3}^{n-1} left(binom{i+1}{4} - binom{i}{4} right)\
& stackrel{binom{3}{4} = 0}{=} & binom{n-1+1}{4} - 0\
& = & binom{n}{4}\
end{eqnarray*}






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    We have to choose four numbers from $[n]$. Assume the largest chosen number is $i+1in[4..n]$. The other three numbers then are arbitrary three numbers in $[i]$.






    share|cite|improve this answer









    $endgroup$





















      3












      $begingroup$

      Let $k=3$ so that $4=k+1$. The binomial coefficient $binom n{k+1}$ counts the $k+1$-element subsets $S$ of the $n$-element set ${0,1,ldots,n-1}$. Let $S$ be such a set and $i=max(S)$. Then $kleq i<n$, and $S=S'cup{i}$ where $S'$ can be any $k$-element subset of the $i$-element set ${0,1,ldots,i-1}$. So
      $$binom n{k+1}=sum_{i=k}^{n-1}binom ik.$$






      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%2f3074465%2fhow-to-prove-binomn4-sum-i-3n-1-binomi3%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









        3












        $begingroup$

        You may transform the sum into a telescoping sum using the "binomial" fact




        • $binom{i}{3} = binom{i+1}{4} - binom{i}{4}$


        Hence,
        begin{eqnarray*} sum_{i=3}^{n-1} binom{i}{3}
        & = & sum_{i=3}^{n-1} left(binom{i+1}{4} - binom{i}{4} right)\
        & stackrel{binom{3}{4} = 0}{=} & binom{n-1+1}{4} - 0\
        & = & binom{n}{4}\
        end{eqnarray*}






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          You may transform the sum into a telescoping sum using the "binomial" fact




          • $binom{i}{3} = binom{i+1}{4} - binom{i}{4}$


          Hence,
          begin{eqnarray*} sum_{i=3}^{n-1} binom{i}{3}
          & = & sum_{i=3}^{n-1} left(binom{i+1}{4} - binom{i}{4} right)\
          & stackrel{binom{3}{4} = 0}{=} & binom{n-1+1}{4} - 0\
          & = & binom{n}{4}\
          end{eqnarray*}






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            You may transform the sum into a telescoping sum using the "binomial" fact




            • $binom{i}{3} = binom{i+1}{4} - binom{i}{4}$


            Hence,
            begin{eqnarray*} sum_{i=3}^{n-1} binom{i}{3}
            & = & sum_{i=3}^{n-1} left(binom{i+1}{4} - binom{i}{4} right)\
            & stackrel{binom{3}{4} = 0}{=} & binom{n-1+1}{4} - 0\
            & = & binom{n}{4}\
            end{eqnarray*}






            share|cite|improve this answer









            $endgroup$



            You may transform the sum into a telescoping sum using the "binomial" fact




            • $binom{i}{3} = binom{i+1}{4} - binom{i}{4}$


            Hence,
            begin{eqnarray*} sum_{i=3}^{n-1} binom{i}{3}
            & = & sum_{i=3}^{n-1} left(binom{i+1}{4} - binom{i}{4} right)\
            & stackrel{binom{3}{4} = 0}{=} & binom{n-1+1}{4} - 0\
            & = & binom{n}{4}\
            end{eqnarray*}







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 15 at 14:15









            trancelocationtrancelocation

            11.9k1825




            11.9k1825























                3












                $begingroup$

                We have to choose four numbers from $[n]$. Assume the largest chosen number is $i+1in[4..n]$. The other three numbers then are arbitrary three numbers in $[i]$.






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  We have to choose four numbers from $[n]$. Assume the largest chosen number is $i+1in[4..n]$. The other three numbers then are arbitrary three numbers in $[i]$.






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    We have to choose four numbers from $[n]$. Assume the largest chosen number is $i+1in[4..n]$. The other three numbers then are arbitrary three numbers in $[i]$.






                    share|cite|improve this answer









                    $endgroup$



                    We have to choose four numbers from $[n]$. Assume the largest chosen number is $i+1in[4..n]$. The other three numbers then are arbitrary three numbers in $[i]$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 15 at 14:33









                    Christian BlatterChristian Blatter

                    174k8115327




                    174k8115327























                        3












                        $begingroup$

                        Let $k=3$ so that $4=k+1$. The binomial coefficient $binom n{k+1}$ counts the $k+1$-element subsets $S$ of the $n$-element set ${0,1,ldots,n-1}$. Let $S$ be such a set and $i=max(S)$. Then $kleq i<n$, and $S=S'cup{i}$ where $S'$ can be any $k$-element subset of the $i$-element set ${0,1,ldots,i-1}$. So
                        $$binom n{k+1}=sum_{i=k}^{n-1}binom ik.$$






                        share|cite|improve this answer









                        $endgroup$


















                          3












                          $begingroup$

                          Let $k=3$ so that $4=k+1$. The binomial coefficient $binom n{k+1}$ counts the $k+1$-element subsets $S$ of the $n$-element set ${0,1,ldots,n-1}$. Let $S$ be such a set and $i=max(S)$. Then $kleq i<n$, and $S=S'cup{i}$ where $S'$ can be any $k$-element subset of the $i$-element set ${0,1,ldots,i-1}$. So
                          $$binom n{k+1}=sum_{i=k}^{n-1}binom ik.$$






                          share|cite|improve this answer









                          $endgroup$
















                            3












                            3








                            3





                            $begingroup$

                            Let $k=3$ so that $4=k+1$. The binomial coefficient $binom n{k+1}$ counts the $k+1$-element subsets $S$ of the $n$-element set ${0,1,ldots,n-1}$. Let $S$ be such a set and $i=max(S)$. Then $kleq i<n$, and $S=S'cup{i}$ where $S'$ can be any $k$-element subset of the $i$-element set ${0,1,ldots,i-1}$. So
                            $$binom n{k+1}=sum_{i=k}^{n-1}binom ik.$$






                            share|cite|improve this answer









                            $endgroup$



                            Let $k=3$ so that $4=k+1$. The binomial coefficient $binom n{k+1}$ counts the $k+1$-element subsets $S$ of the $n$-element set ${0,1,ldots,n-1}$. Let $S$ be such a set and $i=max(S)$. Then $kleq i<n$, and $S=S'cup{i}$ where $S'$ can be any $k$-element subset of the $i$-element set ${0,1,ldots,i-1}$. So
                            $$binom n{k+1}=sum_{i=k}^{n-1}binom ik.$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 15 at 14:34









                            Marc van LeeuwenMarc van Leeuwen

                            87.4k5109224




                            87.4k5109224






























                                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%2f3074465%2fhow-to-prove-binomn4-sum-i-3n-1-binomi3%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

                                Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

                                Image
                                28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
                                Read more

                                Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

                                Image
                                up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
                                Read more

                                A Topological Invariant for $pi_3(U(n))$

                                Image
                                3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
                                Read more