Prove $sum_{k=0}^n binom{m+k}{m} = binom{m+n+1}{m+1}$ using the binomium of Newton












1












$begingroup$


Prove
$$ sum_{k=0}^n binom{m+k}{m} = binom{m+n+1}{m+1} $$
by considering the coefficient of $x^n$ in $(1-x)^{-1} (1-x)^{-m-1} = (1-x)^{-m-2}$.



I have succeeded in proving this using induction and in a combinatorial way, but I need to practice with power series.



Background: third year mathematics at university



My attempt



The given equality can be rewritten as



$$ left( sum_{k=0}^infty binom{-1} k (-x)^k right) left( sum_{ell=0}^infty binom{-m-1}{ell} (-x)^ell right) = sum_{n=0}^infty binom{-m-2}{n} x^n (-1)^n $$



and we can work out the left-hand side as



$$ sum_{n=0}^infty sum_{k+ell=n} binom{-1}{k} binom{-m-1}{ell} (-x)^{k+ell} = sum_{n=0}^infty sum_{k=0}^n binom{-1}{k} binom{-m-1}{n-k} x^n (-1)^n $$



and now we equate the coefficients in front of $x^n$ on both sides to get



$$ sum_{k=0}^n binom{-1}{k} binom{-m-1}{n-k} = binom{-m-2}{n}. $$



But how to proceed? Because I don't see the original equation in this.






combinatorics binomial-coefficients







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Hockey Stick Identity
    $endgroup$
    – Yadati Kiran
    Dec 19 '18 at 8:20










  • $begingroup$
    @YadatiKiran Thanks for the link, it is indeed the same! Now to figure out from the many answers how to prove it using power series.
    $endgroup$
    – PHPirate
    Dec 19 '18 at 8:52










  • $begingroup$
    Ok maybe it is not quite the same, there are ideas in those answers similar to mine but the problem here remains.
    $endgroup$
    – PHPirate
    Dec 19 '18 at 9:10
















1












$begingroup$


Prove
$$ sum_{k=0}^n binom{m+k}{m} = binom{m+n+1}{m+1} $$
by considering the coefficient of $x^n$ in $(1-x)^{-1} (1-x)^{-m-1} = (1-x)^{-m-2}$.



I have succeeded in proving this using induction and in a combinatorial way, but I need to practice with power series.



Background: third year mathematics at university



My attempt



The given equality can be rewritten as



$$ left( sum_{k=0}^infty binom{-1} k (-x)^k right) left( sum_{ell=0}^infty binom{-m-1}{ell} (-x)^ell right) = sum_{n=0}^infty binom{-m-2}{n} x^n (-1)^n $$



and we can work out the left-hand side as



$$ sum_{n=0}^infty sum_{k+ell=n} binom{-1}{k} binom{-m-1}{ell} (-x)^{k+ell} = sum_{n=0}^infty sum_{k=0}^n binom{-1}{k} binom{-m-1}{n-k} x^n (-1)^n $$



and now we equate the coefficients in front of $x^n$ on both sides to get



$$ sum_{k=0}^n binom{-1}{k} binom{-m-1}{n-k} = binom{-m-2}{n}. $$



But how to proceed? Because I don't see the original equation in this.






combinatorics binomial-coefficients







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Hockey Stick Identity
    $endgroup$
    – Yadati Kiran
    Dec 19 '18 at 8:20










  • $begingroup$
    @YadatiKiran Thanks for the link, it is indeed the same! Now to figure out from the many answers how to prove it using power series.
    $endgroup$
    – PHPirate
    Dec 19 '18 at 8:52










  • $begingroup$
    Ok maybe it is not quite the same, there are ideas in those answers similar to mine but the problem here remains.
    $endgroup$
    – PHPirate
    Dec 19 '18 at 9:10














1












1








1





$begingroup$


Prove
$$ sum_{k=0}^n binom{m+k}{m} = binom{m+n+1}{m+1} $$
by considering the coefficient of $x^n$ in $(1-x)^{-1} (1-x)^{-m-1} = (1-x)^{-m-2}$.



I have succeeded in proving this using induction and in a combinatorial way, but I need to practice with power series.



Background: third year mathematics at university



My attempt



The given equality can be rewritten as



$$ left( sum_{k=0}^infty binom{-1} k (-x)^k right) left( sum_{ell=0}^infty binom{-m-1}{ell} (-x)^ell right) = sum_{n=0}^infty binom{-m-2}{n} x^n (-1)^n $$



and we can work out the left-hand side as



$$ sum_{n=0}^infty sum_{k+ell=n} binom{-1}{k} binom{-m-1}{ell} (-x)^{k+ell} = sum_{n=0}^infty sum_{k=0}^n binom{-1}{k} binom{-m-1}{n-k} x^n (-1)^n $$



and now we equate the coefficients in front of $x^n$ on both sides to get



$$ sum_{k=0}^n binom{-1}{k} binom{-m-1}{n-k} = binom{-m-2}{n}. $$



But how to proceed? Because I don't see the original equation in this.






combinatorics binomial-coefficients







share|cite|improve this question









$endgroup$




Prove
$$ sum_{k=0}^n binom{m+k}{m} = binom{m+n+1}{m+1} $$
by considering the coefficient of $x^n$ in $(1-x)^{-1} (1-x)^{-m-1} = (1-x)^{-m-2}$.



I have succeeded in proving this using induction and in a combinatorial way, but I need to practice with power series.



Background: third year mathematics at university



My attempt



The given equality can be rewritten as



$$ left( sum_{k=0}^infty binom{-1} k (-x)^k right) left( sum_{ell=0}^infty binom{-m-1}{ell} (-x)^ell right) = sum_{n=0}^infty binom{-m-2}{n} x^n (-1)^n $$



and we can work out the left-hand side as



$$ sum_{n=0}^infty sum_{k+ell=n} binom{-1}{k} binom{-m-1}{ell} (-x)^{k+ell} = sum_{n=0}^infty sum_{k=0}^n binom{-1}{k} binom{-m-1}{n-k} x^n (-1)^n $$



and now we equate the coefficients in front of $x^n$ on both sides to get



$$ sum_{k=0}^n binom{-1}{k} binom{-m-1}{n-k} = binom{-m-2}{n}. $$



But how to proceed? Because I don't see the original equation in this.






combinatorics binomial-coefficients




combinatorics binomial-coefficients






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 19 '18 at 8:06









PHPiratePHPirate

328417




328417








  • 1




    $begingroup$
    Hockey Stick Identity
    $endgroup$
    – Yadati Kiran
    Dec 19 '18 at 8:20










  • $begingroup$
    @YadatiKiran Thanks for the link, it is indeed the same! Now to figure out from the many answers how to prove it using power series.
    $endgroup$
    – PHPirate
    Dec 19 '18 at 8:52










  • $begingroup$
    Ok maybe it is not quite the same, there are ideas in those answers similar to mine but the problem here remains.
    $endgroup$
    – PHPirate
    Dec 19 '18 at 9:10














  • 1




    $begingroup$
    Hockey Stick Identity
    $endgroup$
    – Yadati Kiran
    Dec 19 '18 at 8:20










  • $begingroup$
    @YadatiKiran Thanks for the link, it is indeed the same! Now to figure out from the many answers how to prove it using power series.
    $endgroup$
    – PHPirate
    Dec 19 '18 at 8:52










  • $begingroup$
    Ok maybe it is not quite the same, there are ideas in those answers similar to mine but the problem here remains.
    $endgroup$
    – PHPirate
    Dec 19 '18 at 9:10








1




1




$begingroup$
Hockey Stick Identity
$endgroup$
– Yadati Kiran
Dec 19 '18 at 8:20




$begingroup$
Hockey Stick Identity
$endgroup$
– Yadati Kiran
Dec 19 '18 at 8:20












$begingroup$
@YadatiKiran Thanks for the link, it is indeed the same! Now to figure out from the many answers how to prove it using power series.
$endgroup$
– PHPirate
Dec 19 '18 at 8:52




$begingroup$
@YadatiKiran Thanks for the link, it is indeed the same! Now to figure out from the many answers how to prove it using power series.
$endgroup$
– PHPirate
Dec 19 '18 at 8:52












$begingroup$
Ok maybe it is not quite the same, there are ideas in those answers similar to mine but the problem here remains.
$endgroup$
– PHPirate
Dec 19 '18 at 9:10




$begingroup$
Ok maybe it is not quite the same, there are ideas in those answers similar to mine but the problem here remains.
$endgroup$
– PHPirate
Dec 19 '18 at 9:10










1 Answer
1






active

oldest

votes


















1












$begingroup$

I mean, you're pretty much done from where you got at. Just note that ${ -1 choose k}=(-1)^k$ and ${-m-2 choose n}= {m+n+1 choose n}$(up to some minus sign) and ${-m-1 choose k}={m+k choose k}$(use this instead of ${-m-1 choose n-k}$ on the LHS of your identity by just changing k to n-k). To prove these identities just write out your binomial coefficients in full.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Indeed it almost works out, I have the LHS equal but for the RHS I have $binom{-m-2}{n}=(-1)^n binom{m+n+1}{n}$, so it is almost equal to $binom{m+n+1}{m+1}$ except for the minus sign, which stays, right?
    $endgroup$
    – PHPirate
    Dec 19 '18 at 9:36










  • $begingroup$
    What do you mean by the "minus sign stays"? The minus signs on the LHS should balance those on the RHS, please check carefully.
    $endgroup$
    – Sorin Tirc
    Dec 19 '18 at 9:39










  • $begingroup$
    For the LHS I have that they are exactly equal: $binom{-1}{k} binom{-m-1}{k} = (-1)^k (-1)^k binom{m+k}{k}=binom{m+k}{m}$ as you say, so that's not where the $(-1)^n$ disappears.
    $endgroup$
    – PHPirate
    Dec 19 '18 at 9:40










  • $begingroup$
    Oh I see, I have a $(-1)^n$ wrong in my attempt in the question! Thanks!
    $endgroup$
    – PHPirate
    Dec 19 '18 at 9:45











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

I mean, you're pretty much done from where you got at. Just note that ${ -1 choose k}=(-1)^k$ and ${-m-2 choose n}= {m+n+1 choose n}$(up to some minus sign) and ${-m-1 choose k}={m+k choose k}$(use this instead of ${-m-1 choose n-k}$ on the LHS of your identity by just changing k to n-k). To prove these identities just write out your binomial coefficients in full.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Indeed it almost works out, I have the LHS equal but for the RHS I have $binom{-m-2}{n}=(-1)^n binom{m+n+1}{n}$, so it is almost equal to $binom{m+n+1}{m+1}$ except for the minus sign, which stays, right?
    $endgroup$
    – PHPirate
    Dec 19 '18 at 9:36










  • $begingroup$
    What do you mean by the "minus sign stays"? The minus signs on the LHS should balance those on the RHS, please check carefully.
    $endgroup$
    – Sorin Tirc
    Dec 19 '18 at 9:39










  • $begingroup$
    For the LHS I have that they are exactly equal: $binom{-1}{k} binom{-m-1}{k} = (-1)^k (-1)^k binom{m+k}{k}=binom{m+k}{m}$ as you say, so that's not where the $(-1)^n$ disappears.
    $endgroup$
    – PHPirate
    Dec 19 '18 at 9:40










  • $begingroup$
    Oh I see, I have a $(-1)^n$ wrong in my attempt in the question! Thanks!
    $endgroup$
    – PHPirate
    Dec 19 '18 at 9:45
















1












$begingroup$

I mean, you're pretty much done from where you got at. Just note that ${ -1 choose k}=(-1)^k$ and ${-m-2 choose n}= {m+n+1 choose n}$(up to some minus sign) and ${-m-1 choose k}={m+k choose k}$(use this instead of ${-m-1 choose n-k}$ on the LHS of your identity by just changing k to n-k). To prove these identities just write out your binomial coefficients in full.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Indeed it almost works out, I have the LHS equal but for the RHS I have $binom{-m-2}{n}=(-1)^n binom{m+n+1}{n}$, so it is almost equal to $binom{m+n+1}{m+1}$ except for the minus sign, which stays, right?
    $endgroup$
    – PHPirate
    Dec 19 '18 at 9:36










  • $begingroup$
    What do you mean by the "minus sign stays"? The minus signs on the LHS should balance those on the RHS, please check carefully.
    $endgroup$
    – Sorin Tirc
    Dec 19 '18 at 9:39










  • $begingroup$
    For the LHS I have that they are exactly equal: $binom{-1}{k} binom{-m-1}{k} = (-1)^k (-1)^k binom{m+k}{k}=binom{m+k}{m}$ as you say, so that's not where the $(-1)^n$ disappears.
    $endgroup$
    – PHPirate
    Dec 19 '18 at 9:40










  • $begingroup$
    Oh I see, I have a $(-1)^n$ wrong in my attempt in the question! Thanks!
    $endgroup$
    – PHPirate
    Dec 19 '18 at 9:45














1












1








1





$begingroup$

I mean, you're pretty much done from where you got at. Just note that ${ -1 choose k}=(-1)^k$ and ${-m-2 choose n}= {m+n+1 choose n}$(up to some minus sign) and ${-m-1 choose k}={m+k choose k}$(use this instead of ${-m-1 choose n-k}$ on the LHS of your identity by just changing k to n-k). To prove these identities just write out your binomial coefficients in full.






share|cite|improve this answer









$endgroup$



I mean, you're pretty much done from where you got at. Just note that ${ -1 choose k}=(-1)^k$ and ${-m-2 choose n}= {m+n+1 choose n}$(up to some minus sign) and ${-m-1 choose k}={m+k choose k}$(use this instead of ${-m-1 choose n-k}$ on the LHS of your identity by just changing k to n-k). To prove these identities just write out your binomial coefficients in full.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 19 '18 at 9:14









Sorin TircSorin Tirc

1,675213




1,675213












  • $begingroup$
    Indeed it almost works out, I have the LHS equal but for the RHS I have $binom{-m-2}{n}=(-1)^n binom{m+n+1}{n}$, so it is almost equal to $binom{m+n+1}{m+1}$ except for the minus sign, which stays, right?
    $endgroup$
    – PHPirate
    Dec 19 '18 at 9:36










  • $begingroup$
    What do you mean by the "minus sign stays"? The minus signs on the LHS should balance those on the RHS, please check carefully.
    $endgroup$
    – Sorin Tirc
    Dec 19 '18 at 9:39










  • $begingroup$
    For the LHS I have that they are exactly equal: $binom{-1}{k} binom{-m-1}{k} = (-1)^k (-1)^k binom{m+k}{k}=binom{m+k}{m}$ as you say, so that's not where the $(-1)^n$ disappears.
    $endgroup$
    – PHPirate
    Dec 19 '18 at 9:40










  • $begingroup$
    Oh I see, I have a $(-1)^n$ wrong in my attempt in the question! Thanks!
    $endgroup$
    – PHPirate
    Dec 19 '18 at 9:45


















  • $begingroup$
    Indeed it almost works out, I have the LHS equal but for the RHS I have $binom{-m-2}{n}=(-1)^n binom{m+n+1}{n}$, so it is almost equal to $binom{m+n+1}{m+1}$ except for the minus sign, which stays, right?
    $endgroup$
    – PHPirate
    Dec 19 '18 at 9:36










  • $begingroup$
    What do you mean by the "minus sign stays"? The minus signs on the LHS should balance those on the RHS, please check carefully.
    $endgroup$
    – Sorin Tirc
    Dec 19 '18 at 9:39










  • $begingroup$
    For the LHS I have that they are exactly equal: $binom{-1}{k} binom{-m-1}{k} = (-1)^k (-1)^k binom{m+k}{k}=binom{m+k}{m}$ as you say, so that's not where the $(-1)^n$ disappears.
    $endgroup$
    – PHPirate
    Dec 19 '18 at 9:40










  • $begingroup$
    Oh I see, I have a $(-1)^n$ wrong in my attempt in the question! Thanks!
    $endgroup$
    – PHPirate
    Dec 19 '18 at 9:45
















$begingroup$
Indeed it almost works out, I have the LHS equal but for the RHS I have $binom{-m-2}{n}=(-1)^n binom{m+n+1}{n}$, so it is almost equal to $binom{m+n+1}{m+1}$ except for the minus sign, which stays, right?
$endgroup$
– PHPirate
Dec 19 '18 at 9:36




$begingroup$
Indeed it almost works out, I have the LHS equal but for the RHS I have $binom{-m-2}{n}=(-1)^n binom{m+n+1}{n}$, so it is almost equal to $binom{m+n+1}{m+1}$ except for the minus sign, which stays, right?
$endgroup$
– PHPirate
Dec 19 '18 at 9:36












$begingroup$
What do you mean by the "minus sign stays"? The minus signs on the LHS should balance those on the RHS, please check carefully.
$endgroup$
– Sorin Tirc
Dec 19 '18 at 9:39




$begingroup$
What do you mean by the "minus sign stays"? The minus signs on the LHS should balance those on the RHS, please check carefully.
$endgroup$
– Sorin Tirc
Dec 19 '18 at 9:39












$begingroup$
For the LHS I have that they are exactly equal: $binom{-1}{k} binom{-m-1}{k} = (-1)^k (-1)^k binom{m+k}{k}=binom{m+k}{m}$ as you say, so that's not where the $(-1)^n$ disappears.
$endgroup$
– PHPirate
Dec 19 '18 at 9:40




$begingroup$
For the LHS I have that they are exactly equal: $binom{-1}{k} binom{-m-1}{k} = (-1)^k (-1)^k binom{m+k}{k}=binom{m+k}{m}$ as you say, so that's not where the $(-1)^n$ disappears.
$endgroup$
– PHPirate
Dec 19 '18 at 9:40












$begingroup$
Oh I see, I have a $(-1)^n$ wrong in my attempt in the question! Thanks!
$endgroup$
– PHPirate
Dec 19 '18 at 9:45




$begingroup$
Oh I see, I have a $(-1)^n$ wrong in my attempt in the question! Thanks!
$endgroup$
– PHPirate
Dec 19 '18 at 9:45


















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