Proof for this binomial coefficient's equation











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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







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  • 1




    Possible duplicate of Prove $sumlimits_{i=0}^nbinom{i+k-1}{k-1}=binom{n+k}{k}$ (a.k.a. Hockey-Stick Identity)
    – user10354138
    17 hours ago










  • @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...)
    – Robert Z
    16 hours ago










  • @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.
    – user10354138
    14 hours ago










  • @user10354138 Fine. So we have a different opinion on this matter. Have a nice day.
    – Robert Z
    12 hours ago















up vote
4
down vote

favorite
3













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









New contributor




るまし is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 1




    Possible duplicate of Prove $sumlimits_{i=0}^nbinom{i+k-1}{k-1}=binom{n+k}{k}$ (a.k.a. Hockey-Stick Identity)
    – user10354138
    17 hours ago










  • @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...)
    – Robert Z
    16 hours ago










  • @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.
    – user10354138
    14 hours ago










  • @user10354138 Fine. So we have a different opinion on this matter. Have a nice day.
    – Robert Z
    12 hours ago













up vote
4
down vote

favorite
3









up vote
4
down vote

favorite
3






3






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









New contributor




るまし is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












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









New contributor




るまし is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









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るまし is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this question




share|cite|improve this question








edited 14 hours ago









Martin Sleziak

44.6k7115269




44.6k7115269






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asked 18 hours ago









るまし

235




235




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  • 1




    Possible duplicate of Prove $sumlimits_{i=0}^nbinom{i+k-1}{k-1}=binom{n+k}{k}$ (a.k.a. Hockey-Stick Identity)
    – user10354138
    17 hours ago










  • @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...)
    – Robert Z
    16 hours ago










  • @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.
    – user10354138
    14 hours ago










  • @user10354138 Fine. So we have a different opinion on this matter. Have a nice day.
    – Robert Z
    12 hours ago














  • 1




    Possible duplicate of Prove $sumlimits_{i=0}^nbinom{i+k-1}{k-1}=binom{n+k}{k}$ (a.k.a. Hockey-Stick Identity)
    – user10354138
    17 hours ago










  • @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...)
    – Robert Z
    16 hours ago










  • @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.
    – user10354138
    14 hours ago










  • @user10354138 Fine. So we have a different opinion on this matter. Have a nice day.
    – Robert Z
    12 hours ago








1




1




Possible duplicate of Prove $sumlimits_{i=0}^nbinom{i+k-1}{k-1}=binom{n+k}{k}$ (a.k.a. Hockey-Stick Identity)
– user10354138
17 hours ago




Possible duplicate of Prove $sumlimits_{i=0}^nbinom{i+k-1}{k-1}=binom{n+k}{k}$ (a.k.a. Hockey-Stick Identity)
– user10354138
17 hours ago












@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...)
– Robert Z
16 hours ago




@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...)
– Robert Z
16 hours ago












@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.
– user10354138
14 hours ago




@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.
– user10354138
14 hours ago












@user10354138 Fine. So we have a different opinion on this matter. Have a nice day.
– Robert Z
12 hours ago




@user10354138 Fine. So we have a different opinion on this matter. Have a nice day.
– Robert Z
12 hours ago










2 Answers
2






active

oldest

votes

















up vote
3
down vote



accepted










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






























    up vote
    5
    down vote













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



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



    1. Hockey-Stick Identity






    share|cite|improve this answer























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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      3
      down vote



      accepted










      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



























        up vote
        3
        down vote



        accepted










        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

























          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          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














          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 17 hours ago

























          answered 17 hours ago









          Robert Z

          91.9k1058129




          91.9k1058129






















              up vote
              5
              down vote













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



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



              1. Hockey-Stick Identity






              share|cite|improve this answer



























                up vote
                5
                down vote













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



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



                1. Hockey-Stick Identity






                share|cite|improve this answer

























                  up vote
                  5
                  down vote










                  up vote
                  5
                  down vote









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



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



                  1. Hockey-Stick Identity






                  share|cite|improve this answer














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



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



                  1. Hockey-Stick Identity







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 17 hours ago

























                  answered 17 hours ago









                  Anubhab Ghosal

                  73012




                  73012






















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