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
$begingroup$
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
asked Mar 22 at 3:15
BenBen
61
$endgroup$
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
|
show 2 more comments
$begingroup$
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
asked Mar 22 at 3:15
BenBen
61
$endgroup$
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
$begingroup$
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
|
show 2 more comments
$begingroup$
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
asked Mar 22 at 3:15
BenBen
61
$endgroup$
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
asked Mar 22 at 3:15
BenBen
61
asked Mar 22 at 3:15
BenBen
61
asked Mar 22 at 3:15
BenBen
61
asked Mar 22 at 3:15
BenBen
61
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
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
$begingroup$
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
|
show 2 more comments
$begingroup$
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
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
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
answered Mar 22 at 5:40
Felix MarinFelix Marin
68.9k7110147
$endgroup$
$begingroup$
The color#ffdis 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
add a comment |
$begingroup$
We have
$$sum_q=0^n m+qchoose q = sum_qge 0 m+qchoose q [[0le qle n]]
\ = sum_qge 0 m+qchoose q [z^n] fracz^q1-z
= [z^n] frac11-z sum_qge 0 m+qchoose q z^q
\= [z^n] frac11-z frac1(1-z)^m+1
= [z^n] frac1(1-z)^m+2
= n+m+1choose n.$$
answered Mar 22 at 14:38
Marko RiedelMarko Riedel
41.2k340111
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
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
answered Mar 22 at 5:40
Felix MarinFelix Marin
68.9k7110147
$endgroup$
$begingroup$
The color#ffdis 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
add a comment |
$begingroup$
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
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
answered Mar 22 at 5:40
Felix MarinFelix Marin
68.9k7110147
$endgroup$
$begingroup$
The color#ffdis 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
add a comment |
$begingroup$
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
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
answered Mar 22 at 5:40
Felix MarinFelix Marin
68.9k7110147
$endgroup$
$newcommandbbx[1],bbox[15px,border:1px groove navy]displaystyle#1,
newcommandbraces[1]leftlbrace,#1,rightrbrace
newcommandbracks[1]leftlbrack,#1,rightrbrack
newcommandddmathrmd
newcommandds[1]displaystyle#1
newcommandexpo[1],mathrme^#1,
newcommandicmathrmi
newcommandmc[1]mathcal#1
newcommandmrm[1]mathrm#1
newcommandpars[1]left(,#1,right)
newcommandpartiald[3][]fracpartial^#1 #2partial #3^#1
newcommandroot[2][],sqrt[#1],#2,,
newcommandtotald[3][]fracmathrmd^#1 #2mathrmd #3^#1
newcommandverts[1]leftvert,#1,rightvert$
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
answered Mar 22 at 5:40
Felix MarinFelix Marin
68.9k7110147
answered Mar 22 at 5:40
Felix MarinFelix Marin
68.9k7110147
answered Mar 22 at 5:40
Felix MarinFelix Marin
68.9k7110147
answered Mar 22 at 5:40
Felix MarinFelix Marin
68.9k7110147
68.9k7110147
$begingroup$
The color#ffdis 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
add a comment |
$begingroup$
The color#ffdis 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
add a comment |
$begingroup$
We have
$$sum_q=0^n m+qchoose q = sum_qge 0 m+qchoose q [[0le qle n]]
\ = sum_qge 0 m+qchoose q [z^n] fracz^q1-z
= [z^n] frac11-z sum_qge 0 m+qchoose q z^q
\= [z^n] frac11-z frac1(1-z)^m+1
= [z^n] frac1(1-z)^m+2
= n+m+1choose n.$$
answered Mar 22 at 14:38
Marko RiedelMarko Riedel
41.2k340111
$endgroup$
add a comment |
$begingroup$
We have
$$sum_q=0^n m+qchoose q = sum_qge 0 m+qchoose q [[0le qle n]]
\ = sum_qge 0 m+qchoose q [z^n] fracz^q1-z
= [z^n] frac11-z sum_qge 0 m+qchoose q z^q
\= [z^n] frac11-z frac1(1-z)^m+1
= [z^n] frac1(1-z)^m+2
= n+m+1choose n.$$
answered Mar 22 at 14:38
Marko RiedelMarko Riedel
41.2k340111
$endgroup$
add a comment |
$begingroup$
We have
$$sum_q=0^n m+qchoose q = sum_qge 0 m+qchoose q [[0le qle n]]
\ = sum_qge 0 m+qchoose q [z^n] fracz^q1-z
= [z^n] frac11-z sum_qge 0 m+qchoose q z^q
\= [z^n] frac11-z frac1(1-z)^m+1
= [z^n] frac1(1-z)^m+2
= n+m+1choose n.$$
answered Mar 22 at 14:38
Marko RiedelMarko Riedel
41.2k340111
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We have
$$sum_q=0^n m+qchoose q = sum_qge 0 m+qchoose q [[0le qle n]]
\ = sum_qge 0 m+qchoose q [z^n] fracz^q1-z
= [z^n] frac11-z sum_qge 0 m+qchoose q z^q
\= [z^n] frac11-z frac1(1-z)^m+1
= [z^n] frac1(1-z)^m+2
= n+m+1choose n.$$
answered Mar 22 at 14:38
Marko RiedelMarko Riedel
41.2k340111
answered Mar 22 at 14:38
Marko RiedelMarko Riedel
41.2k340111
answered Mar 22 at 14:38
Marko RiedelMarko Riedel
41.2k340111
answered Mar 22 at 14:38
Marko RiedelMarko Riedel
41.2k340111
41.2k340111
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You may want to see this: math.stackexchange.com/questions/1784613/…. (Replace their $r$ with your $n$ and their $n$ with your $m+1$.)
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– Minus One-Twelfth
Mar 22 at 3:17
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This is easy to show by induction ... try it ?
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– Donald Splutterwit
Mar 22 at 3:22
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Alternative answer ... Yes someone can explain it !
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– Donald Splutterwit
Mar 22 at 3:23
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@DonaldSplutterwit I know that it's done by induction but still sure how. So yes, I've tried it.
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– Ben
Mar 22 at 4:05
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@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?
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– Ben
Mar 22 at 4:06