Prove $sum_i=0^nbinomm+ii=binomm+n+1n$ (another Hockey-Stick Identity?) [closed]What is the story behind $n+1 choose k = n choose k + n choose k-1$?Another Hockey Stick IdentityIdentity involving binomial coefficients: $sum_j=0^i (-1)^i+jbinom n-ji-jbinom mj=sum_j=0^i (-1)^i+jbinom n-j+ki-jbinom m+kj$Prove $sumlimits_i=0^nbinomi+k-1k-1=binomn+kk$ (a.k.a. Hockey-Stick Identity)Proof of the Hockey-Stick Identity: $sumlimits_t=0^n binom tk = binomn+1k+1$Prove the identity $sum^n_k=0binomm+kk = binomn+m+1n$Hockey-Stick Theorem for Multinomial CoefficientsBinomial identity $sum_i=0^n-1-jbinomni+jbinomi+jj(-1)^i=binomnj(-1)^n+j+1$Another Hockey Stick IdentityEnumerative interpretation of generalized $q$-hockey stick identityHint on how to prove the identity $sum_0 leq k leq n binomkm = binomn + 1m + 1$Proving Pascal's Triangle and Hockey-Stick Identity using Combinatorics

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Prove $sum_i=0^nbinomm+ii=binomm+n+1n$ (another Hockey-Stick Identity?) [closed]


What is the story behind $n+1 choose k = n choose k + n choose k-1$?Another Hockey Stick IdentityIdentity involving binomial coefficients: $sum_j=0^i (-1)^i+jbinom n-ji-jbinom mj=sum_j=0^i (-1)^i+jbinom n-j+ki-jbinom m+kj$Prove $sumlimits_i=0^nbinomi+k-1k-1=binomn+kk$ (a.k.a. Hockey-Stick Identity)Proof of the Hockey-Stick Identity: $sumlimits_t=0^n binom tk = binomn+1k+1$Prove the identity $sum^n_k=0binomm+kk = binomn+m+1n$Hockey-Stick Theorem for Multinomial CoefficientsBinomial identity $sum_i=0^n-1-jbinomni+jbinomi+jj(-1)^i=binomnj(-1)^n+j+1$Another Hockey Stick IdentityEnumerative interpretation of generalized $q$-hockey stick identityHint on how to prove the identity $sum_0 leq k leq n binomkm = binomn + 1m + 1$Proving Pascal's Triangle and Hockey-Stick Identity using Combinatorics













1












$begingroup$


Let $n$ be a nonnegative integer, and $m$ a positive integer. Could someone explain to me why the identity
$$
sum_i=0^nbinomm+ii=binomm+n+1n
$$
holds?






discrete-mathematics binomial-coefficients







share|cite|improve this question









$endgroup$



closed as off-topic by Hanul Jeon, Leucippus, José Carlos Santos, Eevee Trainer, Parcly Taxel Mar 26 at 3:27


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Hanul Jeon, Leucippus, José Carlos Santos, Eevee Trainer, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.















  • $begingroup$
    You may want to see this: math.stackexchange.com/questions/1784613/…. (Replace their $r$ with your $n$ and their $n$ with your $m+1$.)
    $endgroup$
    – Minus One-Twelfth
    Mar 22 at 3:17











  • $begingroup$
    This is easy to show by induction ... try it ?
    $endgroup$
    – Donald Splutterwit
    Mar 22 at 3:22










  • $begingroup$
    Alternative answer ... Yes someone can explain it !
    $endgroup$
    – Donald Splutterwit
    Mar 22 at 3:23










  • $begingroup$
    @DonaldSplutterwit I know that it's done by induction but still sure how. So yes, I've tried it.
    $endgroup$
    – Ben
    Mar 22 at 4:05










  • $begingroup$
    @MinusOne-Twelfth It's almost impossible to keep the terms straight but I substituted my stuff over to the best reply on that post but it still isn't very clear. Do you know the induction method?
    $endgroup$
    – Ben
    Mar 22 at 4:06















1












$begingroup$


Let $n$ be a nonnegative integer, and $m$ a positive integer. Could someone explain to me why the identity
$$
sum_i=0^nbinomm+ii=binomm+n+1n
$$
holds?






discrete-mathematics binomial-coefficients







share|cite|improve this question









$endgroup$



closed as off-topic by Hanul Jeon, Leucippus, José Carlos Santos, Eevee Trainer, Parcly Taxel Mar 26 at 3:27


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Hanul Jeon, Leucippus, José Carlos Santos, Eevee Trainer, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.















  • $begingroup$
    You may want to see this: math.stackexchange.com/questions/1784613/…. (Replace their $r$ with your $n$ and their $n$ with your $m+1$.)
    $endgroup$
    – Minus One-Twelfth
    Mar 22 at 3:17











  • $begingroup$
    This is easy to show by induction ... try it ?
    $endgroup$
    – Donald Splutterwit
    Mar 22 at 3:22










  • $begingroup$
    Alternative answer ... Yes someone can explain it !
    $endgroup$
    – Donald Splutterwit
    Mar 22 at 3:23










  • $begingroup$
    @DonaldSplutterwit I know that it's done by induction but still sure how. So yes, I've tried it.
    $endgroup$
    – Ben
    Mar 22 at 4:05










  • $begingroup$
    @MinusOne-Twelfth It's almost impossible to keep the terms straight but I substituted my stuff over to the best reply on that post but it still isn't very clear. Do you know the induction method?
    $endgroup$
    – Ben
    Mar 22 at 4:06













1












1








1





$begingroup$


Let $n$ be a nonnegative integer, and $m$ a positive integer. Could someone explain to me why the identity
$$
sum_i=0^nbinomm+ii=binomm+n+1n
$$
holds?






discrete-mathematics binomial-coefficients







share|cite|improve this question









$endgroup$




Let $n$ be a nonnegative integer, and $m$ a positive integer. Could someone explain to me why the identity
$$
sum_i=0^nbinomm+ii=binomm+n+1n
$$
holds?






discrete-mathematics binomial-coefficients




discrete-mathematics binomial-coefficients






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 22 at 3:15









BenBen

61




61




closed as off-topic by Hanul Jeon, Leucippus, José Carlos Santos, Eevee Trainer, Parcly Taxel Mar 26 at 3:27


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Hanul Jeon, Leucippus, José Carlos Santos, Eevee Trainer, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Hanul Jeon, Leucippus, José Carlos Santos, Eevee Trainer, Parcly Taxel Mar 26 at 3:27


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Hanul Jeon, Leucippus, José Carlos Santos, Eevee Trainer, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.











  • $begingroup$
    You may want to see this: math.stackexchange.com/questions/1784613/…. (Replace their $r$ with your $n$ and their $n$ with your $m+1$.)
    $endgroup$
    – Minus One-Twelfth
    Mar 22 at 3:17











  • $begingroup$
    This is easy to show by induction ... try it ?
    $endgroup$
    – Donald Splutterwit
    Mar 22 at 3:22










  • $begingroup$
    Alternative answer ... Yes someone can explain it !
    $endgroup$
    – Donald Splutterwit
    Mar 22 at 3:23










  • $begingroup$
    @DonaldSplutterwit I know that it's done by induction but still sure how. So yes, I've tried it.
    $endgroup$
    – Ben
    Mar 22 at 4:05










  • $begingroup$
    @MinusOne-Twelfth It's almost impossible to keep the terms straight but I substituted my stuff over to the best reply on that post but it still isn't very clear. Do you know the induction method?
    $endgroup$
    – Ben
    Mar 22 at 4:06
















  • $begingroup$
    You may want to see this: math.stackexchange.com/questions/1784613/…. (Replace their $r$ with your $n$ and their $n$ with your $m+1$.)
    $endgroup$
    – Minus One-Twelfth
    Mar 22 at 3:17











  • $begingroup$
    This is easy to show by induction ... try it ?
    $endgroup$
    – Donald Splutterwit
    Mar 22 at 3:22










  • $begingroup$
    Alternative answer ... Yes someone can explain it !
    $endgroup$
    – Donald Splutterwit
    Mar 22 at 3:23










  • $begingroup$
    @DonaldSplutterwit I know that it's done by induction but still sure how. So yes, I've tried it.
    $endgroup$
    – Ben
    Mar 22 at 4:05










  • $begingroup$
    @MinusOne-Twelfth It's almost impossible to keep the terms straight but I substituted my stuff over to the best reply on that post but it still isn't very clear. Do you know the induction method?
    $endgroup$
    – Ben
    Mar 22 at 4:06















$begingroup$
You may want to see this: math.stackexchange.com/questions/1784613/…. (Replace their $r$ with your $n$ and their $n$ with your $m+1$.)
$endgroup$
– Minus One-Twelfth
Mar 22 at 3:17





$begingroup$
You may want to see this: math.stackexchange.com/questions/1784613/…. (Replace their $r$ with your $n$ and their $n$ with your $m+1$.)
$endgroup$
– Minus One-Twelfth
Mar 22 at 3:17













$begingroup$
This is easy to show by induction ... try it ?
$endgroup$
– Donald Splutterwit
Mar 22 at 3:22




$begingroup$
This is easy to show by induction ... try it ?
$endgroup$
– Donald Splutterwit
Mar 22 at 3:22












$begingroup$
Alternative answer ... Yes someone can explain it !
$endgroup$
– Donald Splutterwit
Mar 22 at 3:23




$begingroup$
Alternative answer ... Yes someone can explain it !
$endgroup$
– Donald Splutterwit
Mar 22 at 3:23












$begingroup$
@DonaldSplutterwit I know that it's done by induction but still sure how. So yes, I've tried it.
$endgroup$
– Ben
Mar 22 at 4:05




$begingroup$
@DonaldSplutterwit I know that it's done by induction but still sure how. So yes, I've tried it.
$endgroup$
– Ben
Mar 22 at 4:05












$begingroup$
@MinusOne-Twelfth It's almost impossible to keep the terms straight but I substituted my stuff over to the best reply on that post but it still isn't very clear. Do you know the induction method?
$endgroup$
– Ben
Mar 22 at 4:06




$begingroup$
@MinusOne-Twelfth It's almost impossible to keep the terms straight but I substituted my stuff over to the best reply on that post but it still isn't very clear. Do you know the induction method?
$endgroup$
– Ben
Mar 22 at 4:06










2 Answers
2






active

oldest

votes


















2












$begingroup$

$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
beginalign
&bbox[10px,#ffd]sum_i = 0^nm + i choose i =
sum_i = 0^npars-1^i-m -i + i - 1 choose i
\[5mm] = &
sum_i = 0^npars-1^i-m - 1 choose i =
sum_i = 0^npars-1^ibracksz^ipars1 + z^-m - 1
\[5mm] = &
bracksz^0pars1 + z^-m - 1
sum_i = 0^npars-,1 over z^i
=
bracksz^0pars1 + z^-m - 1,
pars-1/z^n + 1 - 1 over pars-1/z - 1
\[5mm] = &
bracksz^0pars1 + z^-m - 1,
pars-1^n + 1 - z^n + 1 over -1 - z
,z over z^n + 1
\[5mm] = &
bracksz^npars1 + z^-m - 2,
bracksz^n + 1 - pars-1^n + 1 =
pars-1^nbracksz^npars1 + z^-m - 2
\[5mm] = &
pars-1^n-m - 2 choose n =
pars-1^nbracksm + 2 + n - 1choose npars-1^n
\[5mm] = & bbxm + n + 1 choose n
endalign






share|cite|improve this answer









$endgroup$












  • $begingroup$
    The color #ffd is so ease to work with :)
    $endgroup$
    – Le Anh Dung
    Mar 22 at 5:52






  • 2




    $begingroup$
    @LeAnhDung It clearly shows the very beginning while the box shows the very end.
    $endgroup$
    – Felix Marin
    Mar 22 at 5:54


















2












$begingroup$

We have



$$sum_q=0^n m+qchoose q = sum_qge 0 m+qchoose q [[0le qle n]]
\ = sum_qge 0 m+qchoose q [z^n] fracz^q1-z
= [z^n] frac11-z sum_qge 0 m+qchoose q z^q
\= [z^n] frac11-z frac1(1-z)^m+1
= [z^n] frac1(1-z)^m+2
= n+m+1choose n.$$






share|cite|improve this answer









$endgroup$



















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    $newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
    newcommandbraces[1]leftlbrace,#1,rightrbrace
    newcommandbracks[1]leftlbrack,#1,rightrbrack
    newcommandddmathrmd
    newcommandds[1]displaystyle#1
    newcommandexpo[1],mathrme^#1,
    newcommandicmathrmi
    newcommandmc[1]mathcal#1
    newcommandmrm[1]mathrm#1
    newcommandpars[1]left(,#1,right)
    newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
    newcommandroot[2][],sqrt[#1],#2,,
    newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
    newcommandverts[1]leftvert,#1,rightvert$
    beginalign
    &bbox[10px,#ffd]sum_i = 0^nm + i choose i =
    sum_i = 0^npars-1^i-m -i + i - 1 choose i
    \[5mm] = &
    sum_i = 0^npars-1^i-m - 1 choose i =
    sum_i = 0^npars-1^ibracksz^ipars1 + z^-m - 1
    \[5mm] = &
    bracksz^0pars1 + z^-m - 1
    sum_i = 0^npars-,1 over z^i
    =
    bracksz^0pars1 + z^-m - 1,
    pars-1/z^n + 1 - 1 over pars-1/z - 1
    \[5mm] = &
    bracksz^0pars1 + z^-m - 1,
    pars-1^n + 1 - z^n + 1 over -1 - z
    ,z over z^n + 1
    \[5mm] = &
    bracksz^npars1 + z^-m - 2,
    bracksz^n + 1 - pars-1^n + 1 =
    pars-1^nbracksz^npars1 + z^-m - 2
    \[5mm] = &
    pars-1^n-m - 2 choose n =
    pars-1^nbracksm + 2 + n - 1choose npars-1^n
    \[5mm] = & bbxm + n + 1 choose n
    endalign






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      The color #ffd is so ease to work with :)
      $endgroup$
      – Le Anh Dung
      Mar 22 at 5:52






    • 2




      $begingroup$
      @LeAnhDung It clearly shows the very beginning while the box shows the very end.
      $endgroup$
      – Felix Marin
      Mar 22 at 5:54















    2












    $begingroup$

    $newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
    newcommandbraces[1]leftlbrace,#1,rightrbrace
    newcommandbracks[1]leftlbrack,#1,rightrbrack
    newcommandddmathrmd
    newcommandds[1]displaystyle#1
    newcommandexpo[1],mathrme^#1,
    newcommandicmathrmi
    newcommandmc[1]mathcal#1
    newcommandmrm[1]mathrm#1
    newcommandpars[1]left(,#1,right)
    newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
    newcommandroot[2][],sqrt[#1],#2,,
    newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
    newcommandverts[1]leftvert,#1,rightvert$
    beginalign
    &bbox[10px,#ffd]sum_i = 0^nm + i choose i =
    sum_i = 0^npars-1^i-m -i + i - 1 choose i
    \[5mm] = &
    sum_i = 0^npars-1^i-m - 1 choose i =
    sum_i = 0^npars-1^ibracksz^ipars1 + z^-m - 1
    \[5mm] = &
    bracksz^0pars1 + z^-m - 1
    sum_i = 0^npars-,1 over z^i
    =
    bracksz^0pars1 + z^-m - 1,
    pars-1/z^n + 1 - 1 over pars-1/z - 1
    \[5mm] = &
    bracksz^0pars1 + z^-m - 1,
    pars-1^n + 1 - z^n + 1 over -1 - z
    ,z over z^n + 1
    \[5mm] = &
    bracksz^npars1 + z^-m - 2,
    bracksz^n + 1 - pars-1^n + 1 =
    pars-1^nbracksz^npars1 + z^-m - 2
    \[5mm] = &
    pars-1^n-m - 2 choose n =
    pars-1^nbracksm + 2 + n - 1choose npars-1^n
    \[5mm] = & bbxm + n + 1 choose n
    endalign






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      The color #ffd is so ease to work with :)
      $endgroup$
      – Le Anh Dung
      Mar 22 at 5:52






    • 2




      $begingroup$
      @LeAnhDung It clearly shows the very beginning while the box shows the very end.
      $endgroup$
      – Felix Marin
      Mar 22 at 5:54













    2












    2








    2





    $begingroup$

    $newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
    newcommandbraces[1]leftlbrace,#1,rightrbrace
    newcommandbracks[1]leftlbrack,#1,rightrbrack
    newcommandddmathrmd
    newcommandds[1]displaystyle#1
    newcommandexpo[1],mathrme^#1,
    newcommandicmathrmi
    newcommandmc[1]mathcal#1
    newcommandmrm[1]mathrm#1
    newcommandpars[1]left(,#1,right)
    newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
    newcommandroot[2][],sqrt[#1],#2,,
    newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
    newcommandverts[1]leftvert,#1,rightvert$
    beginalign
    &bbox[10px,#ffd]sum_i = 0^nm + i choose i =
    sum_i = 0^npars-1^i-m -i + i - 1 choose i
    \[5mm] = &
    sum_i = 0^npars-1^i-m - 1 choose i =
    sum_i = 0^npars-1^ibracksz^ipars1 + z^-m - 1
    \[5mm] = &
    bracksz^0pars1 + z^-m - 1
    sum_i = 0^npars-,1 over z^i
    =
    bracksz^0pars1 + z^-m - 1,
    pars-1/z^n + 1 - 1 over pars-1/z - 1
    \[5mm] = &
    bracksz^0pars1 + z^-m - 1,
    pars-1^n + 1 - z^n + 1 over -1 - z
    ,z over z^n + 1
    \[5mm] = &
    bracksz^npars1 + z^-m - 2,
    bracksz^n + 1 - pars-1^n + 1 =
    pars-1^nbracksz^npars1 + z^-m - 2
    \[5mm] = &
    pars-1^n-m - 2 choose n =
    pars-1^nbracksm + 2 + n - 1choose npars-1^n
    \[5mm] = & bbxm + n + 1 choose n
    endalign






    share|cite|improve this answer









    $endgroup$



    $newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
    newcommandbraces[1]leftlbrace,#1,rightrbrace
    newcommandbracks[1]leftlbrack,#1,rightrbrack
    newcommandddmathrmd
    newcommandds[1]displaystyle#1
    newcommandexpo[1],mathrme^#1,
    newcommandicmathrmi
    newcommandmc[1]mathcal#1
    newcommandmrm[1]mathrm#1
    newcommandpars[1]left(,#1,right)
    newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
    newcommandroot[2][],sqrt[#1],#2,,
    newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
    newcommandverts[1]leftvert,#1,rightvert$
    beginalign
    &bbox[10px,#ffd]sum_i = 0^nm + i choose i =
    sum_i = 0^npars-1^i-m -i + i - 1 choose i
    \[5mm] = &
    sum_i = 0^npars-1^i-m - 1 choose i =
    sum_i = 0^npars-1^ibracksz^ipars1 + z^-m - 1
    \[5mm] = &
    bracksz^0pars1 + z^-m - 1
    sum_i = 0^npars-,1 over z^i
    =
    bracksz^0pars1 + z^-m - 1,
    pars-1/z^n + 1 - 1 over pars-1/z - 1
    \[5mm] = &
    bracksz^0pars1 + z^-m - 1,
    pars-1^n + 1 - z^n + 1 over -1 - z
    ,z over z^n + 1
    \[5mm] = &
    bracksz^npars1 + z^-m - 2,
    bracksz^n + 1 - pars-1^n + 1 =
    pars-1^nbracksz^npars1 + z^-m - 2
    \[5mm] = &
    pars-1^n-m - 2 choose n =
    pars-1^nbracksm + 2 + n - 1choose npars-1^n
    \[5mm] = & bbxm + n + 1 choose n
    endalign







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Mar 22 at 5:40









    Felix MarinFelix Marin

    68.9k7110147




    68.9k7110147











    • $begingroup$
      The color #ffd is so ease to work with :)
      $endgroup$
      – Le Anh Dung
      Mar 22 at 5:52






    • 2




      $begingroup$
      @LeAnhDung It clearly shows the very beginning while the box shows the very end.
      $endgroup$
      – Felix Marin
      Mar 22 at 5:54
















    • $begingroup$
      The color #ffd is so ease to work with :)
      $endgroup$
      – Le Anh Dung
      Mar 22 at 5:52






    • 2




      $begingroup$
      @LeAnhDung It clearly shows the very beginning while the box shows the very end.
      $endgroup$
      – Felix Marin
      Mar 22 at 5:54















    $begingroup$
    The color #ffd is so ease to work with :)
    $endgroup$
    – Le Anh Dung
    Mar 22 at 5:52




    $begingroup$
    The color #ffd is so ease to work with :)
    $endgroup$
    – Le Anh Dung
    Mar 22 at 5:52




    2




    2




    $begingroup$
    @LeAnhDung It clearly shows the very beginning while the box shows the very end.
    $endgroup$
    – Felix Marin
    Mar 22 at 5:54




    $begingroup$
    @LeAnhDung It clearly shows the very beginning while the box shows the very end.
    $endgroup$
    – Felix Marin
    Mar 22 at 5:54











    2












    $begingroup$

    We have



    $$sum_q=0^n m+qchoose q = sum_qge 0 m+qchoose q [[0le qle n]]
    \ = sum_qge 0 m+qchoose q [z^n] fracz^q1-z
    = [z^n] frac11-z sum_qge 0 m+qchoose q z^q
    \= [z^n] frac11-z frac1(1-z)^m+1
    = [z^n] frac1(1-z)^m+2
    = n+m+1choose n.$$






    share|cite|improve this answer









    $endgroup$

















      2












      $begingroup$

      We have



      $$sum_q=0^n m+qchoose q = sum_qge 0 m+qchoose q [[0le qle n]]
      \ = sum_qge 0 m+qchoose q [z^n] fracz^q1-z
      = [z^n] frac11-z sum_qge 0 m+qchoose q z^q
      \= [z^n] frac11-z frac1(1-z)^m+1
      = [z^n] frac1(1-z)^m+2
      = n+m+1choose n.$$






      share|cite|improve this answer









      $endgroup$















        2












        2








        2





        $begingroup$

        We have



        $$sum_q=0^n m+qchoose q = sum_qge 0 m+qchoose q [[0le qle n]]
        \ = sum_qge 0 m+qchoose q [z^n] fracz^q1-z
        = [z^n] frac11-z sum_qge 0 m+qchoose q z^q
        \= [z^n] frac11-z frac1(1-z)^m+1
        = [z^n] frac1(1-z)^m+2
        = n+m+1choose n.$$






        share|cite|improve this answer









        $endgroup$



        We have



        $$sum_q=0^n m+qchoose q = sum_qge 0 m+qchoose q [[0le qle n]]
        \ = sum_qge 0 m+qchoose q [z^n] fracz^q1-z
        = [z^n] frac11-z sum_qge 0 m+qchoose q z^q
        \= [z^n] frac11-z frac1(1-z)^m+1
        = [z^n] frac1(1-z)^m+2
        = n+m+1choose n.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 22 at 14:38









        Marko RiedelMarko Riedel

        41.2k340111




        41.2k340111













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