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












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$begingroup$


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







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  • $begingroup$
    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













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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

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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.






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    $begingroup$

    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






























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