Proof for this binomial coefficient's equation












4












$begingroup$



For $k, l in mathbb N$
$$sum_{i=0}^ksum_{j=0}^lbinom{i+j}i=binom{k+l+2}{k+1}-1$$
How can I prove this?




I thought some ideas with Pascal's triangle, counting paths on the grid and simple deformation of the formula.



It can be checked here (wolframalpha).



If the proof is difficult, please let me know the main idea.



Sorry for my poor English.



Thank you.



EDIT:
I got the great and short proof using Hockey-stick identity by Anubhab Ghosal, but because of this form, I could also get the Robert Z's specialized answer.
Then I don't think it is fully duplicate.






combinatorics summation binomial-coefficients







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Possible duplicate of Prove $sumlimits_{i=0}^nbinom{i+k-1}{k-1}=binom{n+k}{k}$ (a.k.a. Hockey-Stick Identity)
    $endgroup$
    – user10354138
    Dec 12 '18 at 10:07










  • $begingroup$
    @user10354138 The double sum is related Hockey-Stick Identity but it is not a duplicate of the linked question. Moreover OP asks for a combinatorial proof (counting paths...)
    $endgroup$
    – Robert Z
    Dec 12 '18 at 11:30










  • $begingroup$
    @RobertZ The OP didn't ask for a combinatorial proof, only for a proof ("How can I prove this?" in the quote, not "How can I prove this combinatorially?"). It can be proved by applying the linked result twice, so it is a duplicate.
    $endgroup$
    – user10354138
    Dec 12 '18 at 13:18










  • $begingroup$
    @user10354138 Fine. So we have a different opinion on this matter. Have a nice day.
    $endgroup$
    – Robert Z
    Dec 12 '18 at 15:48
















4












$begingroup$



For $k, l in mathbb N$
$$sum_{i=0}^ksum_{j=0}^lbinom{i+j}i=binom{k+l+2}{k+1}-1$$
How can I prove this?




I thought some ideas with Pascal's triangle, counting paths on the grid and simple deformation of the formula.



It can be checked here (wolframalpha).



If the proof is difficult, please let me know the main idea.



Sorry for my poor English.



Thank you.



EDIT:
I got the great and short proof using Hockey-stick identity by Anubhab Ghosal, but because of this form, I could also get the Robert Z's specialized answer.
Then I don't think it is fully duplicate.






combinatorics summation binomial-coefficients







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Possible duplicate of Prove $sumlimits_{i=0}^nbinom{i+k-1}{k-1}=binom{n+k}{k}$ (a.k.a. Hockey-Stick Identity)
    $endgroup$
    – user10354138
    Dec 12 '18 at 10:07










  • $begingroup$
    @user10354138 The double sum is related Hockey-Stick Identity but it is not a duplicate of the linked question. Moreover OP asks for a combinatorial proof (counting paths...)
    $endgroup$
    – Robert Z
    Dec 12 '18 at 11:30










  • $begingroup$
    @RobertZ The OP didn't ask for a combinatorial proof, only for a proof ("How can I prove this?" in the quote, not "How can I prove this combinatorially?"). It can be proved by applying the linked result twice, so it is a duplicate.
    $endgroup$
    – user10354138
    Dec 12 '18 at 13:18










  • $begingroup$
    @user10354138 Fine. So we have a different opinion on this matter. Have a nice day.
    $endgroup$
    – Robert Z
    Dec 12 '18 at 15:48














4












4








4


2



$begingroup$



For $k, l in mathbb N$
$$sum_{i=0}^ksum_{j=0}^lbinom{i+j}i=binom{k+l+2}{k+1}-1$$
How can I prove this?




I thought some ideas with Pascal's triangle, counting paths on the grid and simple deformation of the formula.



It can be checked here (wolframalpha).



If the proof is difficult, please let me know the main idea.



Sorry for my poor English.



Thank you.



EDIT:
I got the great and short proof using Hockey-stick identity by Anubhab Ghosal, but because of this form, I could also get the Robert Z's specialized answer.
Then I don't think it is fully duplicate.






combinatorics summation binomial-coefficients







share|cite|improve this question











$endgroup$





For $k, l in mathbb N$
$$sum_{i=0}^ksum_{j=0}^lbinom{i+j}i=binom{k+l+2}{k+1}-1$$
How can I prove this?




I thought some ideas with Pascal's triangle, counting paths on the grid and simple deformation of the formula.



It can be checked here (wolframalpha).



If the proof is difficult, please let me know the main idea.



Sorry for my poor English.



Thank you.



EDIT:
I got the great and short proof using Hockey-stick identity by Anubhab Ghosal, but because of this form, I could also get the Robert Z's specialized answer.
Then I don't think it is fully duplicate.






combinatorics summation binomial-coefficients




combinatorics summation binomial-coefficients






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 13:54









Martin Sleziak

44.9k10122277




44.9k10122277










asked Dec 12 '18 at 9:58









るましるまし

235




235








  • 1




    $begingroup$
    Possible duplicate of Prove $sumlimits_{i=0}^nbinom{i+k-1}{k-1}=binom{n+k}{k}$ (a.k.a. Hockey-Stick Identity)
    $endgroup$
    – user10354138
    Dec 12 '18 at 10:07










  • $begingroup$
    @user10354138 The double sum is related Hockey-Stick Identity but it is not a duplicate of the linked question. Moreover OP asks for a combinatorial proof (counting paths...)
    $endgroup$
    – Robert Z
    Dec 12 '18 at 11:30










  • $begingroup$
    @RobertZ The OP didn't ask for a combinatorial proof, only for a proof ("How can I prove this?" in the quote, not "How can I prove this combinatorially?"). It can be proved by applying the linked result twice, so it is a duplicate.
    $endgroup$
    – user10354138
    Dec 12 '18 at 13:18










  • $begingroup$
    @user10354138 Fine. So we have a different opinion on this matter. Have a nice day.
    $endgroup$
    – Robert Z
    Dec 12 '18 at 15:48














  • 1




    $begingroup$
    Possible duplicate of Prove $sumlimits_{i=0}^nbinom{i+k-1}{k-1}=binom{n+k}{k}$ (a.k.a. Hockey-Stick Identity)
    $endgroup$
    – user10354138
    Dec 12 '18 at 10:07










  • $begingroup$
    @user10354138 The double sum is related Hockey-Stick Identity but it is not a duplicate of the linked question. Moreover OP asks for a combinatorial proof (counting paths...)
    $endgroup$
    – Robert Z
    Dec 12 '18 at 11:30










  • $begingroup$
    @RobertZ The OP didn't ask for a combinatorial proof, only for a proof ("How can I prove this?" in the quote, not "How can I prove this combinatorially?"). It can be proved by applying the linked result twice, so it is a duplicate.
    $endgroup$
    – user10354138
    Dec 12 '18 at 13:18










  • $begingroup$
    @user10354138 Fine. So we have a different opinion on this matter. Have a nice day.
    $endgroup$
    – Robert Z
    Dec 12 '18 at 15:48








1




1




$begingroup$
Possible duplicate of Prove $sumlimits_{i=0}^nbinom{i+k-1}{k-1}=binom{n+k}{k}$ (a.k.a. Hockey-Stick Identity)
$endgroup$
– user10354138
Dec 12 '18 at 10:07




$begingroup$
Possible duplicate of Prove $sumlimits_{i=0}^nbinom{i+k-1}{k-1}=binom{n+k}{k}$ (a.k.a. Hockey-Stick Identity)
$endgroup$
– user10354138
Dec 12 '18 at 10:07












$begingroup$
@user10354138 The double sum is related Hockey-Stick Identity but it is not a duplicate of the linked question. Moreover OP asks for a combinatorial proof (counting paths...)
$endgroup$
– Robert Z
Dec 12 '18 at 11:30




$begingroup$
@user10354138 The double sum is related Hockey-Stick Identity but it is not a duplicate of the linked question. Moreover OP asks for a combinatorial proof (counting paths...)
$endgroup$
– Robert Z
Dec 12 '18 at 11:30












$begingroup$
@RobertZ The OP didn't ask for a combinatorial proof, only for a proof ("How can I prove this?" in the quote, not "How can I prove this combinatorially?"). It can be proved by applying the linked result twice, so it is a duplicate.
$endgroup$
– user10354138
Dec 12 '18 at 13:18




$begingroup$
@RobertZ The OP didn't ask for a combinatorial proof, only for a proof ("How can I prove this?" in the quote, not "How can I prove this combinatorially?"). It can be proved by applying the linked result twice, so it is a duplicate.
$endgroup$
– user10354138
Dec 12 '18 at 13:18












$begingroup$
@user10354138 Fine. So we have a different opinion on this matter. Have a nice day.
$endgroup$
– Robert Z
Dec 12 '18 at 15:48




$begingroup$
@user10354138 Fine. So we have a different opinion on this matter. Have a nice day.
$endgroup$
– Robert Z
Dec 12 '18 at 15:48










3 Answers
3






active

oldest

votes


















3












$begingroup$

Your idea about a combinatorial proof which is related to counting paths in a grid is a good one!



The binomial $binom{i+j}i$
counts the paths on the grid from $(0,0)$ to $(i,j)$ moving only right or up. So the double sum
$$sum_{i=0}^ksum_{j=0}^lbinom{i+j}i-1$$
counts the number of all such paths from $(0,0)$ to any vertex inside the rectangle $(0,k)times (0,l)$ different from $(0,0)$.



Now consider the paths from $(0,0)$ to $(k+1,l+1)$ different from $$(0,0)to (0,l+1)to (k+1,l+1)quadtext{and}quad
(0,0)to (k+1,0)to (k+1,l+1)$$ which are
$$binom{k+l+2}{k+1}-2.$$



Now any path of the first kind can be completed to a path of the second kind by changing direction, going to the boundary of the rectangle
$(0,k+1)times(0,l+1)$ and then moving to the corner $(k+1,l+1)$ along the side.



Is this a bijection between the first set of paths and the second one?






share|cite|improve this answer











$endgroup$





















    5












    $begingroup$

    $$displaystylesum_{i=0}^ksum_{j=0}^lbinom{i+j}i=sum_{i=0}^ksum_{j=i}^{i+l}binom{j}i=sum_{i=0}^kbinom{i+l+1}{i+1} ^{[1]}$$



    $$=sum_{i=0}^kbinom{i+l+1}{l}=sum_{i=l}^{k+l+1}binom{i}{l}−1=binom{k+l+2}{k+1}-1 ^{[1]}$$



    1. Hockey-Stick Identity






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
      newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
      newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
      newcommand{dd}{mathrm{d}}
      newcommand{ds}[1]{displaystyle{#1}}
      newcommand{expo}[1]{,mathrm{e}^{#1},}
      newcommand{ic}{mathrm{i}}
      newcommand{mc}[1]{mathcal{#1}}
      newcommand{mrm}[1]{mathrm{#1}}
      newcommand{pars}[1]{left(,{#1},right)}
      newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
      newcommand{root}[2]{,sqrt[#1]{,{#2},},}
      newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
      newcommand{verts}[1]{leftvert,{#1},rightvert}$




      $ds{sum_{i = 0}^{k}sum_{j = 0}^{ell}
      {i + j choose i} = {k + ell + 2 choose k + 1} - 1: {LARGE ?}.qquad k, ell in mathbb{N}}$      .




      begin{align}
      &bbox[10px,#ffd]{sum_{i = 0}^{k}sum_{j = 0}^{ell}
      {i + j choose i}} =
      sum_{i = 0}^{k}sum_{j = 0}^{ell}{i + j choose j} =
      sum_{i = 0}^{k}sum_{j = 0}^{ell}{-i - 1 choose j}
      pars{-1}^{,j}
      \[5mm] = &
      sum_{i = 0}^{k}sum_{j = 0}^{ell}pars{-1}^{,j}
      bracks{z^{, j}}pars{1 + z}^{-i - 1} =
      sum_{i = 0}^{k}sum_{j = 0}^{ell}pars{-1}^{,j}
      bracks{z^{0}}{1 over z^{, j}},pars{1 + z}^{-i - 1}
      \[5mm] = &
      bracks{z^{0}}sum_{i = 0}^{k}pars{1 over 1 + z}^{i + 1}
      sum_{j = 0}^{ell}pars{-,{1 over z}}^{,j}
      \[5mm] = &
      bracks{z^{0}}braces{{1 over 1 + z},
      {bracks{1/pars{1 + z}}^{k + 1} - 1 over 1/pars{1 + z} - 1}}
      braces{{pars{-1/z}^{ell + 1} - 1 over -1/z - 1}}
      \[5mm] = &
      bracks{z^{0}}braces{%
      {1 - pars{1 + z}^{k + 1} over -z}
      ,{1 over pars{1 + z}^{k + 1}}}
      braces{{pars{-1}^{ell + 1} - z^{ell + 1} over -1 - z},{z over z^{ell + 1}}}
      \[5mm] = &
      bracks{z^{ell + 1}}braces{1 - {1 over pars{1 + z}^{k + 1}}}
      braces{z^{ell + 1} + pars{-1}^{ell} over 1 + z}
      \[5mm] = &
      pars{-1}^{ell}bracks{z^{ell + 1}}
      bracks{pars{1 + z}^{-1} - pars{1 + z}^{-k - 2}}
      \[5mm] = &
      pars{-1}^{ell}bracks{pars{-1}^{ell + 1} - {-k - 2 choose ell + 1}}
      \[5mm] = &
      -1 - pars{-1}^{ell},{-bracks{-k - 2} + bracks{ell + 1} - 1 choose ell + 1}pars{-1}^{ell + 1}
      \[5mm] = &
      -1 + { k + ell + 2 choose ell + 1} =
      bbx{{k + ell + 2 choose k + 1} - 1}
      end{align}






      share|cite|improve this answer









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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        Your idea about a combinatorial proof which is related to counting paths in a grid is a good one!



        The binomial $binom{i+j}i$
        counts the paths on the grid from $(0,0)$ to $(i,j)$ moving only right or up. So the double sum
        $$sum_{i=0}^ksum_{j=0}^lbinom{i+j}i-1$$
        counts the number of all such paths from $(0,0)$ to any vertex inside the rectangle $(0,k)times (0,l)$ different from $(0,0)$.



        Now consider the paths from $(0,0)$ to $(k+1,l+1)$ different from $$(0,0)to (0,l+1)to (k+1,l+1)quadtext{and}quad
        (0,0)to (k+1,0)to (k+1,l+1)$$         which are
        $$binom{k+l+2}{k+1}-2.$$



        Now any path of the first kind can be completed to a path of the second kind by changing direction, going to the boundary of the rectangle
        $(0,k+1)times(0,l+1)$ and then moving to the corner $(k+1,l+1)$ along the side.



        Is this a bijection between the first set of paths and the second one?






        share|cite|improve this answer











        $endgroup$


















          3












          $begingroup$

          Your idea about a combinatorial proof which is related to counting paths in a grid is a good one!



          The binomial $binom{i+j}i$
          counts the paths on the grid from $(0,0)$ to $(i,j)$ moving only right or up. So the double sum
          $$sum_{i=0}^ksum_{j=0}^lbinom{i+j}i-1$$
          counts the number of all such paths from $(0,0)$ to any vertex inside the rectangle $(0,k)times (0,l)$ different from $(0,0)$.



          Now consider the paths from $(0,0)$ to $(k+1,l+1)$ different from $$(0,0)to (0,l+1)to (k+1,l+1)quadtext{and}quad
          (0,0)to (k+1,0)to (k+1,l+1)$$           which are
          $$binom{k+l+2}{k+1}-2.$$



          Now any path of the first kind can be completed to a path of the second kind by changing direction, going to the boundary of the rectangle
          $(0,k+1)times(0,l+1)$ and then moving to the corner $(k+1,l+1)$ along the side.



          Is this a bijection between the first set of paths and the second one?






          share|cite|improve this answer











          $endgroup$
















            3












            3








            3





            $begingroup$

            Your idea about a combinatorial proof which is related to counting paths in a grid is a good one!



            The binomial $binom{i+j}i$
            counts the paths on the grid from $(0,0)$ to $(i,j)$ moving only right or up. So the double sum
            $$sum_{i=0}^ksum_{j=0}^lbinom{i+j}i-1$$
            counts the number of all such paths from $(0,0)$ to any vertex inside the rectangle $(0,k)times (0,l)$ different from $(0,0)$.



            Now consider the paths from $(0,0)$ to $(k+1,l+1)$ different from $$(0,0)to (0,l+1)to (k+1,l+1)quadtext{and}quad
            (0,0)to (k+1,0)to (k+1,l+1)$$             which are
            $$binom{k+l+2}{k+1}-2.$$



            Now any path of the first kind can be completed to a path of the second kind by changing direction, going to the boundary of the rectangle
            $(0,k+1)times(0,l+1)$ and then moving to the corner $(k+1,l+1)$ along the side.



            Is this a bijection between the first set of paths and the second one?






            share|cite|improve this answer











            $endgroup$



            Your idea about a combinatorial proof which is related to counting paths in a grid is a good one!



            The binomial $binom{i+j}i$
            counts the paths on the grid from $(0,0)$ to $(i,j)$ moving only right or up. So the double sum
            $$sum_{i=0}^ksum_{j=0}^lbinom{i+j}i-1$$
            counts the number of all such paths from $(0,0)$ to any vertex inside the rectangle $(0,k)times (0,l)$ different from $(0,0)$.



            Now consider the paths from $(0,0)$ to $(k+1,l+1)$ different from $$(0,0)to (0,l+1)to (k+1,l+1)quadtext{and}quad
            (0,0)to (k+1,0)to (k+1,l+1)$$             which are
            $$binom{k+l+2}{k+1}-2.$$



            Now any path of the first kind can be completed to a path of the second kind by changing direction, going to the boundary of the rectangle
            $(0,k+1)times(0,l+1)$ and then moving to the corner $(k+1,l+1)$ along the side.



            Is this a bijection between the first set of paths and the second one?







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 12 '18 at 10:42

























            answered Dec 12 '18 at 10:31









            Robert ZRobert Z

            101k1071144




            101k1071144























                5












                $begingroup$

                $$displaystylesum_{i=0}^ksum_{j=0}^lbinom{i+j}i=sum_{i=0}^ksum_{j=i}^{i+l}binom{j}i=sum_{i=0}^kbinom{i+l+1}{i+1} ^{[1]}$$



                $$=sum_{i=0}^kbinom{i+l+1}{l}=sum_{i=l}^{k+l+1}binom{i}{l}−1=binom{k+l+2}{k+1}-1 ^{[1]}$$



                1. Hockey-Stick Identity






                share|cite|improve this answer











                $endgroup$


















                  5












                  $begingroup$

                  $$displaystylesum_{i=0}^ksum_{j=0}^lbinom{i+j}i=sum_{i=0}^ksum_{j=i}^{i+l}binom{j}i=sum_{i=0}^kbinom{i+l+1}{i+1} ^{[1]}$$



                  $$=sum_{i=0}^kbinom{i+l+1}{l}=sum_{i=l}^{k+l+1}binom{i}{l}−1=binom{k+l+2}{k+1}-1 ^{[1]}$$



                  1. Hockey-Stick Identity






                  share|cite|improve this answer











                  $endgroup$
















                    5












                    5








                    5





                    $begingroup$

                    $$displaystylesum_{i=0}^ksum_{j=0}^lbinom{i+j}i=sum_{i=0}^ksum_{j=i}^{i+l}binom{j}i=sum_{i=0}^kbinom{i+l+1}{i+1} ^{[1]}$$



                    $$=sum_{i=0}^kbinom{i+l+1}{l}=sum_{i=l}^{k+l+1}binom{i}{l}−1=binom{k+l+2}{k+1}-1 ^{[1]}$$



                    1. Hockey-Stick Identity






                    share|cite|improve this answer











                    $endgroup$



                    $$displaystylesum_{i=0}^ksum_{j=0}^lbinom{i+j}i=sum_{i=0}^ksum_{j=i}^{i+l}binom{j}i=sum_{i=0}^kbinom{i+l+1}{i+1} ^{[1]}$$



                    $$=sum_{i=0}^kbinom{i+l+1}{l}=sum_{i=l}^{k+l+1}binom{i}{l}−1=binom{k+l+2}{k+1}-1 ^{[1]}$$



                    1. Hockey-Stick Identity







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 12 '18 at 10:17

























                    answered Dec 12 '18 at 10:12









                    Anubhab GhosalAnubhab Ghosal

                    1,23619




                    1,23619























                        0












                        $begingroup$

                        $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
                        newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
                        newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
                        newcommand{dd}{mathrm{d}}
                        newcommand{ds}[1]{displaystyle{#1}}
                        newcommand{expo}[1]{,mathrm{e}^{#1},}
                        newcommand{ic}{mathrm{i}}
                        newcommand{mc}[1]{mathcal{#1}}
                        newcommand{mrm}[1]{mathrm{#1}}
                        newcommand{pars}[1]{left(,{#1},right)}
                        newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
                        newcommand{root}[2]{,sqrt[#1]{,{#2},},}
                        newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
                        newcommand{verts}[1]{leftvert,{#1},rightvert}$




                        $ds{sum_{i = 0}^{k}sum_{j = 0}^{ell}
                        {i + j choose i} = {k + ell + 2 choose k + 1} - 1: {LARGE ?}.qquad k, ell in mathbb{N}}$                        .




                        begin{align}
                        &bbox[10px,#ffd]{sum_{i = 0}^{k}sum_{j = 0}^{ell}
                        {i + j choose i}} =
                        sum_{i = 0}^{k}sum_{j = 0}^{ell}{i + j choose j} =
                        sum_{i = 0}^{k}sum_{j = 0}^{ell}{-i - 1 choose j}
                        pars{-1}^{,j}
                        \[5mm] = &
                        sum_{i = 0}^{k}sum_{j = 0}^{ell}pars{-1}^{,j}
                        bracks{z^{, j}}pars{1 + z}^{-i - 1} =
                        sum_{i = 0}^{k}sum_{j = 0}^{ell}pars{-1}^{,j}
                        bracks{z^{0}}{1 over z^{, j}},pars{1 + z}^{-i - 1}
                        \[5mm] = &
                        bracks{z^{0}}sum_{i = 0}^{k}pars{1 over 1 + z}^{i + 1}
                        sum_{j = 0}^{ell}pars{-,{1 over z}}^{,j}
                        \[5mm] = &
                        bracks{z^{0}}braces{{1 over 1 + z},
                        {bracks{1/pars{1 + z}}^{k + 1} - 1 over 1/pars{1 + z} - 1}}
                        braces{{pars{-1/z}^{ell + 1} - 1 over -1/z - 1}}
                        \[5mm] = &
                        bracks{z^{0}}braces{%
                        {1 - pars{1 + z}^{k + 1} over -z}
                        ,{1 over pars{1 + z}^{k + 1}}}
                        braces{{pars{-1}^{ell + 1} - z^{ell + 1} over -1 - z},{z over z^{ell + 1}}}
                        \[5mm] = &
                        bracks{z^{ell + 1}}braces{1 - {1 over pars{1 + z}^{k + 1}}}
                        braces{z^{ell + 1} + pars{-1}^{ell} over 1 + z}
                        \[5mm] = &
                        pars{-1}^{ell}bracks{z^{ell + 1}}
                        bracks{pars{1 + z}^{-1} - pars{1 + z}^{-k - 2}}
                        \[5mm] = &
                        pars{-1}^{ell}bracks{pars{-1}^{ell + 1} - {-k - 2 choose ell + 1}}
                        \[5mm] = &
                        -1 - pars{-1}^{ell},{-bracks{-k - 2} + bracks{ell + 1} - 1 choose ell + 1}pars{-1}^{ell + 1}
                        \[5mm] = &
                        -1 + { k + ell + 2 choose ell + 1} =
                        bbx{{k + ell + 2 choose k + 1} - 1}
                        end{align}






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
                          newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
                          newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
                          newcommand{dd}{mathrm{d}}
                          newcommand{ds}[1]{displaystyle{#1}}
                          newcommand{expo}[1]{,mathrm{e}^{#1},}
                          newcommand{ic}{mathrm{i}}
                          newcommand{mc}[1]{mathcal{#1}}
                          newcommand{mrm}[1]{mathrm{#1}}
                          newcommand{pars}[1]{left(,{#1},right)}
                          newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
                          newcommand{root}[2]{,sqrt[#1]{,{#2},},}
                          newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
                          newcommand{verts}[1]{leftvert,{#1},rightvert}$




                          $ds{sum_{i = 0}^{k}sum_{j = 0}^{ell}
                          {i + j choose i} = {k + ell + 2 choose k + 1} - 1: {LARGE ?}.qquad k, ell in mathbb{N}}$                          .




                          begin{align}
                          &bbox[10px,#ffd]{sum_{i = 0}^{k}sum_{j = 0}^{ell}
                          {i + j choose i}} =
                          sum_{i = 0}^{k}sum_{j = 0}^{ell}{i + j choose j} =
                          sum_{i = 0}^{k}sum_{j = 0}^{ell}{-i - 1 choose j}
                          pars{-1}^{,j}
                          \[5mm] = &
                          sum_{i = 0}^{k}sum_{j = 0}^{ell}pars{-1}^{,j}
                          bracks{z^{, j}}pars{1 + z}^{-i - 1} =
                          sum_{i = 0}^{k}sum_{j = 0}^{ell}pars{-1}^{,j}
                          bracks{z^{0}}{1 over z^{, j}},pars{1 + z}^{-i - 1}
                          \[5mm] = &
                          bracks{z^{0}}sum_{i = 0}^{k}pars{1 over 1 + z}^{i + 1}
                          sum_{j = 0}^{ell}pars{-,{1 over z}}^{,j}
                          \[5mm] = &
                          bracks{z^{0}}braces{{1 over 1 + z},
                          {bracks{1/pars{1 + z}}^{k + 1} - 1 over 1/pars{1 + z} - 1}}
                          braces{{pars{-1/z}^{ell + 1} - 1 over -1/z - 1}}
                          \[5mm] = &
                          bracks{z^{0}}braces{%
                          {1 - pars{1 + z}^{k + 1} over -z}
                          ,{1 over pars{1 + z}^{k + 1}}}
                          braces{{pars{-1}^{ell + 1} - z^{ell + 1} over -1 - z},{z over z^{ell + 1}}}
                          \[5mm] = &
                          bracks{z^{ell + 1}}braces{1 - {1 over pars{1 + z}^{k + 1}}}
                          braces{z^{ell + 1} + pars{-1}^{ell} over 1 + z}
                          \[5mm] = &
                          pars{-1}^{ell}bracks{z^{ell + 1}}
                          bracks{pars{1 + z}^{-1} - pars{1 + z}^{-k - 2}}
                          \[5mm] = &
                          pars{-1}^{ell}bracks{pars{-1}^{ell + 1} - {-k - 2 choose ell + 1}}
                          \[5mm] = &
                          -1 - pars{-1}^{ell},{-bracks{-k - 2} + bracks{ell + 1} - 1 choose ell + 1}pars{-1}^{ell + 1}
                          \[5mm] = &
                          -1 + { k + ell + 2 choose ell + 1} =
                          bbx{{k + ell + 2 choose k + 1} - 1}
                          end{align}






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
                            newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
                            newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
                            newcommand{dd}{mathrm{d}}
                            newcommand{ds}[1]{displaystyle{#1}}
                            newcommand{expo}[1]{,mathrm{e}^{#1},}
                            newcommand{ic}{mathrm{i}}
                            newcommand{mc}[1]{mathcal{#1}}
                            newcommand{mrm}[1]{mathrm{#1}}
                            newcommand{pars}[1]{left(,{#1},right)}
                            newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
                            newcommand{root}[2]{,sqrt[#1]{,{#2},},}
                            newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
                            newcommand{verts}[1]{leftvert,{#1},rightvert}$




                            $ds{sum_{i = 0}^{k}sum_{j = 0}^{ell}
                            {i + j choose i} = {k + ell + 2 choose k + 1} - 1: {LARGE ?}.qquad k, ell in mathbb{N}}$                            .




                            begin{align}
                            &bbox[10px,#ffd]{sum_{i = 0}^{k}sum_{j = 0}^{ell}
                            {i + j choose i}} =
                            sum_{i = 0}^{k}sum_{j = 0}^{ell}{i + j choose j} =
                            sum_{i = 0}^{k}sum_{j = 0}^{ell}{-i - 1 choose j}
                            pars{-1}^{,j}
                            \[5mm] = &
                            sum_{i = 0}^{k}sum_{j = 0}^{ell}pars{-1}^{,j}
                            bracks{z^{, j}}pars{1 + z}^{-i - 1} =
                            sum_{i = 0}^{k}sum_{j = 0}^{ell}pars{-1}^{,j}
                            bracks{z^{0}}{1 over z^{, j}},pars{1 + z}^{-i - 1}
                            \[5mm] = &
                            bracks{z^{0}}sum_{i = 0}^{k}pars{1 over 1 + z}^{i + 1}
                            sum_{j = 0}^{ell}pars{-,{1 over z}}^{,j}
                            \[5mm] = &
                            bracks{z^{0}}braces{{1 over 1 + z},
                            {bracks{1/pars{1 + z}}^{k + 1} - 1 over 1/pars{1 + z} - 1}}
                            braces{{pars{-1/z}^{ell + 1} - 1 over -1/z - 1}}
                            \[5mm] = &
                            bracks{z^{0}}braces{%
                            {1 - pars{1 + z}^{k + 1} over -z}
                            ,{1 over pars{1 + z}^{k + 1}}}
                            braces{{pars{-1}^{ell + 1} - z^{ell + 1} over -1 - z},{z over z^{ell + 1}}}
                            \[5mm] = &
                            bracks{z^{ell + 1}}braces{1 - {1 over pars{1 + z}^{k + 1}}}
                            braces{z^{ell + 1} + pars{-1}^{ell} over 1 + z}
                            \[5mm] = &
                            pars{-1}^{ell}bracks{z^{ell + 1}}
                            bracks{pars{1 + z}^{-1} - pars{1 + z}^{-k - 2}}
                            \[5mm] = &
                            pars{-1}^{ell}bracks{pars{-1}^{ell + 1} - {-k - 2 choose ell + 1}}
                            \[5mm] = &
                            -1 - pars{-1}^{ell},{-bracks{-k - 2} + bracks{ell + 1} - 1 choose ell + 1}pars{-1}^{ell + 1}
                            \[5mm] = &
                            -1 + { k + ell + 2 choose ell + 1} =
                            bbx{{k + ell + 2 choose k + 1} - 1}
                            end{align}






                            share|cite|improve this answer









                            $endgroup$



                            $newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
                            newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
                            newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
                            newcommand{dd}{mathrm{d}}
                            newcommand{ds}[1]{displaystyle{#1}}
                            newcommand{expo}[1]{,mathrm{e}^{#1},}
                            newcommand{ic}{mathrm{i}}
                            newcommand{mc}[1]{mathcal{#1}}
                            newcommand{mrm}[1]{mathrm{#1}}
                            newcommand{pars}[1]{left(,{#1},right)}
                            newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
                            newcommand{root}[2]{,sqrt[#1]{,{#2},},}
                            newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
                            newcommand{verts}[1]{leftvert,{#1},rightvert}$




                            $ds{sum_{i = 0}^{k}sum_{j = 0}^{ell}
                            {i + j choose i} = {k + ell + 2 choose k + 1} - 1: {LARGE ?}.qquad k, ell in mathbb{N}}$                            .




                            begin{align}
                            &bbox[10px,#ffd]{sum_{i = 0}^{k}sum_{j = 0}^{ell}
                            {i + j choose i}} =
                            sum_{i = 0}^{k}sum_{j = 0}^{ell}{i + j choose j} =
                            sum_{i = 0}^{k}sum_{j = 0}^{ell}{-i - 1 choose j}
                            pars{-1}^{,j}
                            \[5mm] = &
                            sum_{i = 0}^{k}sum_{j = 0}^{ell}pars{-1}^{,j}
                            bracks{z^{, j}}pars{1 + z}^{-i - 1} =
                            sum_{i = 0}^{k}sum_{j = 0}^{ell}pars{-1}^{,j}
                            bracks{z^{0}}{1 over z^{, j}},pars{1 + z}^{-i - 1}
                            \[5mm] = &
                            bracks{z^{0}}sum_{i = 0}^{k}pars{1 over 1 + z}^{i + 1}
                            sum_{j = 0}^{ell}pars{-,{1 over z}}^{,j}
                            \[5mm] = &
                            bracks{z^{0}}braces{{1 over 1 + z},
                            {bracks{1/pars{1 + z}}^{k + 1} - 1 over 1/pars{1 + z} - 1}}
                            braces{{pars{-1/z}^{ell + 1} - 1 over -1/z - 1}}
                            \[5mm] = &
                            bracks{z^{0}}braces{%
                            {1 - pars{1 + z}^{k + 1} over -z}
                            ,{1 over pars{1 + z}^{k + 1}}}
                            braces{{pars{-1}^{ell + 1} - z^{ell + 1} over -1 - z},{z over z^{ell + 1}}}
                            \[5mm] = &
                            bracks{z^{ell + 1}}braces{1 - {1 over pars{1 + z}^{k + 1}}}
                            braces{z^{ell + 1} + pars{-1}^{ell} over 1 + z}
                            \[5mm] = &
                            pars{-1}^{ell}bracks{z^{ell + 1}}
                            bracks{pars{1 + z}^{-1} - pars{1 + z}^{-k - 2}}
                            \[5mm] = &
                            pars{-1}^{ell}bracks{pars{-1}^{ell + 1} - {-k - 2 choose ell + 1}}
                            \[5mm] = &
                            -1 - pars{-1}^{ell},{-bracks{-k - 2} + bracks{ell + 1} - 1 choose ell + 1}pars{-1}^{ell + 1}
                            \[5mm] = &
                            -1 + { k + ell + 2 choose ell + 1} =
                            bbx{{k + ell + 2 choose k + 1} - 1}
                            end{align}







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                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 20 '18 at 22:11









                            Felix MarinFelix Marin

                            68.8k7110146




                            68.8k7110146






























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