Let n belongs to +ve integer and $(1+x+x^2)^n=sum_{r=0}^{2n} {a_rx^r}$ prove that: $a_r=a_{0<r<2n}$












-1












$begingroup$


Let n belongs to +ve integer and $$(1+x+x^2)^n=sum_{r=0}^{2n} {a_rx^r}$$ prove that: $$a_r=a_{2n-1},{0<r<2n}$$ as well as prove that $$sum_{r=0}^{ n-1} a_r=frac{1}{2}(3^n-a_n)$$. I tried to solve this problem by using multinomial theorem and comparing the coefficients but was confused how should I proceed please help me out.






binomial-coefficients binomial-theorem multinomial-coefficients negative-binomial







share|cite|improve this question









$endgroup$












  • $begingroup$
    Not every $a_r$ between $r=1$ and $r=2n-1$ (inclusive) are equal.
    $endgroup$
    – user10354138
    Dec 6 '18 at 15:59












  • $begingroup$
    Should be $a_r = a_{2n-r}$.
    $endgroup$
    – gandalf61
    Dec 6 '18 at 16:01
















-1












$begingroup$


Let n belongs to +ve integer and $$(1+x+x^2)^n=sum_{r=0}^{2n} {a_rx^r}$$ prove that: $$a_r=a_{2n-1},{0<r<2n}$$ as well as prove that $$sum_{r=0}^{ n-1} a_r=frac{1}{2}(3^n-a_n)$$. I tried to solve this problem by using multinomial theorem and comparing the coefficients but was confused how should I proceed please help me out.






binomial-coefficients binomial-theorem multinomial-coefficients negative-binomial







share|cite|improve this question









$endgroup$












  • $begingroup$
    Not every $a_r$ between $r=1$ and $r=2n-1$ (inclusive) are equal.
    $endgroup$
    – user10354138
    Dec 6 '18 at 15:59












  • $begingroup$
    Should be $a_r = a_{2n-r}$.
    $endgroup$
    – gandalf61
    Dec 6 '18 at 16:01














-1












-1








-1





$begingroup$


Let n belongs to +ve integer and $$(1+x+x^2)^n=sum_{r=0}^{2n} {a_rx^r}$$ prove that: $$a_r=a_{2n-1},{0<r<2n}$$ as well as prove that $$sum_{r=0}^{ n-1} a_r=frac{1}{2}(3^n-a_n)$$. I tried to solve this problem by using multinomial theorem and comparing the coefficients but was confused how should I proceed please help me out.






binomial-coefficients binomial-theorem multinomial-coefficients negative-binomial







share|cite|improve this question









$endgroup$




Let n belongs to +ve integer and $$(1+x+x^2)^n=sum_{r=0}^{2n} {a_rx^r}$$ prove that: $$a_r=a_{2n-1},{0<r<2n}$$ as well as prove that $$sum_{r=0}^{ n-1} a_r=frac{1}{2}(3^n-a_n)$$. I tried to solve this problem by using multinomial theorem and comparing the coefficients but was confused how should I proceed please help me out.






binomial-coefficients binomial-theorem multinomial-coefficients negative-binomial




binomial-coefficients binomial-theorem multinomial-coefficients negative-binomial






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 6 '18 at 15:45









Bipul KumarBipul Kumar

355




355












  • $begingroup$
    Not every $a_r$ between $r=1$ and $r=2n-1$ (inclusive) are equal.
    $endgroup$
    – user10354138
    Dec 6 '18 at 15:59












  • $begingroup$
    Should be $a_r = a_{2n-r}$.
    $endgroup$
    – gandalf61
    Dec 6 '18 at 16:01


















  • $begingroup$
    Not every $a_r$ between $r=1$ and $r=2n-1$ (inclusive) are equal.
    $endgroup$
    – user10354138
    Dec 6 '18 at 15:59












  • $begingroup$
    Should be $a_r = a_{2n-r}$.
    $endgroup$
    – gandalf61
    Dec 6 '18 at 16:01
















$begingroup$
Not every $a_r$ between $r=1$ and $r=2n-1$ (inclusive) are equal.
$endgroup$
– user10354138
Dec 6 '18 at 15:59






$begingroup$
Not every $a_r$ between $r=1$ and $r=2n-1$ (inclusive) are equal.
$endgroup$
– user10354138
Dec 6 '18 at 15:59














$begingroup$
Should be $a_r = a_{2n-r}$.
$endgroup$
– gandalf61
Dec 6 '18 at 16:01




$begingroup$
Should be $a_r = a_{2n-r}$.
$endgroup$
– gandalf61
Dec 6 '18 at 16:01










1 Answer
1






active

oldest

votes


















2












$begingroup$

The first property $a_r=a_{2n-r}$ comes from the symmetry of the original expression. More formally:



$(1+x+x^2)^n=x^{2n}(1+x^{-1}+x^{-2})^n = x^{2n}sum_{r=0}^{2n} {a_rx^{-r}} = sum_{r=0}^{2n} {a_rx^{2n-r}}=sum_{r=0}^{2n} {a_{2n-r}x^r}$



$Rightarrow sum_{r=0}^{2n} {a_rx^r}=sum_{r=0}^{2n} {a_{2n-r}x^r}$



Equating the coefficients of each power of $x$ gives the required result.



Now that we know the coeffciients $a_r$ are symmetric we can rewrite the original expression as:



$(1+x+x^2)^n=sum_{r=0}^{n-1} {a_r(x^r+x^{2n-r}})+a_nx^n$



Note that we split out the middle term $a_nx^n$ because it is not paired with another symmetric term. Now substitute $x=1$ and re-arrange.






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%2f3028655%2flet-n-belongs-to-ve-integer-and-1xx2n-sum-r-02n-a-rxr-prove-th%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    The first property $a_r=a_{2n-r}$ comes from the symmetry of the original expression. More formally:



    $(1+x+x^2)^n=x^{2n}(1+x^{-1}+x^{-2})^n = x^{2n}sum_{r=0}^{2n} {a_rx^{-r}} = sum_{r=0}^{2n} {a_rx^{2n-r}}=sum_{r=0}^{2n} {a_{2n-r}x^r}$



    $Rightarrow sum_{r=0}^{2n} {a_rx^r}=sum_{r=0}^{2n} {a_{2n-r}x^r}$



    Equating the coefficients of each power of $x$ gives the required result.



    Now that we know the coeffciients $a_r$ are symmetric we can rewrite the original expression as:



    $(1+x+x^2)^n=sum_{r=0}^{n-1} {a_r(x^r+x^{2n-r}})+a_nx^n$



    Note that we split out the middle term $a_nx^n$ because it is not paired with another symmetric term. Now substitute $x=1$ and re-arrange.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      The first property $a_r=a_{2n-r}$ comes from the symmetry of the original expression. More formally:



      $(1+x+x^2)^n=x^{2n}(1+x^{-1}+x^{-2})^n = x^{2n}sum_{r=0}^{2n} {a_rx^{-r}} = sum_{r=0}^{2n} {a_rx^{2n-r}}=sum_{r=0}^{2n} {a_{2n-r}x^r}$



      $Rightarrow sum_{r=0}^{2n} {a_rx^r}=sum_{r=0}^{2n} {a_{2n-r}x^r}$



      Equating the coefficients of each power of $x$ gives the required result.



      Now that we know the coeffciients $a_r$ are symmetric we can rewrite the original expression as:



      $(1+x+x^2)^n=sum_{r=0}^{n-1} {a_r(x^r+x^{2n-r}})+a_nx^n$



      Note that we split out the middle term $a_nx^n$ because it is not paired with another symmetric term. Now substitute $x=1$ and re-arrange.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        The first property $a_r=a_{2n-r}$ comes from the symmetry of the original expression. More formally:



        $(1+x+x^2)^n=x^{2n}(1+x^{-1}+x^{-2})^n = x^{2n}sum_{r=0}^{2n} {a_rx^{-r}} = sum_{r=0}^{2n} {a_rx^{2n-r}}=sum_{r=0}^{2n} {a_{2n-r}x^r}$



        $Rightarrow sum_{r=0}^{2n} {a_rx^r}=sum_{r=0}^{2n} {a_{2n-r}x^r}$



        Equating the coefficients of each power of $x$ gives the required result.



        Now that we know the coeffciients $a_r$ are symmetric we can rewrite the original expression as:



        $(1+x+x^2)^n=sum_{r=0}^{n-1} {a_r(x^r+x^{2n-r}})+a_nx^n$



        Note that we split out the middle term $a_nx^n$ because it is not paired with another symmetric term. Now substitute $x=1$ and re-arrange.






        share|cite|improve this answer









        $endgroup$



        The first property $a_r=a_{2n-r}$ comes from the symmetry of the original expression. More formally:



        $(1+x+x^2)^n=x^{2n}(1+x^{-1}+x^{-2})^n = x^{2n}sum_{r=0}^{2n} {a_rx^{-r}} = sum_{r=0}^{2n} {a_rx^{2n-r}}=sum_{r=0}^{2n} {a_{2n-r}x^r}$



        $Rightarrow sum_{r=0}^{2n} {a_rx^r}=sum_{r=0}^{2n} {a_{2n-r}x^r}$



        Equating the coefficients of each power of $x$ gives the required result.



        Now that we know the coeffciients $a_r$ are symmetric we can rewrite the original expression as:



        $(1+x+x^2)^n=sum_{r=0}^{n-1} {a_r(x^r+x^{2n-r}})+a_nx^n$



        Note that we split out the middle term $a_nx^n$ because it is not paired with another symmetric term. Now substitute $x=1$ and re-arrange.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 6 '18 at 16:00









        gandalf61gandalf61

        8,329625




        8,329625






























            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%2f3028655%2flet-n-belongs-to-ve-integer-and-1xx2n-sum-r-02n-a-rxr-prove-th%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

            Quarter-circle Tiles

            Image
            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
            Read more

            build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

            Image
            0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
            Read more

            Mont Emei

            Image
            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
            Read more