Binomial coefficients sumSum of two random variables ( negative binomial distribution )Binomial Theorem identities, evaluate the sumCompact form of sum (binomial coefficients)Sum of binomial distributed random variablesSum of Series of Binomial Coefficients.Sum of product of squared binomial coefficientsInverting binomial coefficientsEvaluating sum of binomial coefficientsConvergence of sum of reciprocals of binomial coefficientsIs there a closed form for the sum of the cubes of the binomial coefficients?Proving an identity involving the alternating sum of products of binomial coefficients
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Binomial coefficients sum
Sum of two random variables ( negative binomial distribution )Binomial Theorem identities, evaluate the sumCompact form of sum (binomial coefficients)Sum of binomial distributed random variablesSum of Series of Binomial Coefficients.Sum of product of squared binomial coefficientsInverting binomial coefficientsEvaluating sum of binomial coefficientsConvergence of sum of reciprocals of binomial coefficientsIs there a closed form for the sum of the cubes of the binomial coefficients?Proving an identity involving the alternating sum of products of binomial coefficients
$begingroup$
Let $n geq 1$ and $N geq 1$ be integers. I am interested in the sum $$sum_k=0^N binomk + n-1n - 1$$ I don't know how to tackle this. I've tried using the definition of $binomnk$ but did not get anywhere.
Could anyone suggest a method of attack for evaluating this sum?
summation binomial-coefficients
asked Mar 28 at 16:24
the manthe man
831716
$endgroup$
add a comment |
$begingroup$
Let $n geq 1$ and $N geq 1$ be integers. I am interested in the sum $$sum_k=0^N binomk + n-1n - 1$$ I don't know how to tackle this. I've tried using the definition of $binomnk$ but did not get anywhere.
Could anyone suggest a method of attack for evaluating this sum?
summation binomial-coefficients
asked Mar 28 at 16:24
the manthe man
831716
$endgroup$
1
$begingroup$
You can use the result of the user drhab below and proof e.g. by induction.
$endgroup$
– user90369
Mar 28 at 16:53
add a comment |
$begingroup$
Let $n geq 1$ and $N geq 1$ be integers. I am interested in the sum $$sum_k=0^N binomk + n-1n - 1$$ I don't know how to tackle this. I've tried using the definition of $binomnk$ but did not get anywhere.
Could anyone suggest a method of attack for evaluating this sum?
summation binomial-coefficients
asked Mar 28 at 16:24
the manthe man
831716
$endgroup$
Let $n geq 1$ and $N geq 1$ be integers. I am interested in the sum $$sum_k=0^N binomk + n-1n - 1$$ I don't know how to tackle this. I've tried using the definition of $binomnk$ but did not get anywhere.
Could anyone suggest a method of attack for evaluating this sum?
summation binomial-coefficients
summation binomial-coefficients
asked Mar 28 at 16:24
the manthe man
831716
asked Mar 28 at 16:24
the manthe man
831716
asked Mar 28 at 16:24
the manthe man
831716
asked Mar 28 at 16:24
the manthe man
831716
asked Mar 28 at 16:24
the manthe man
831716
831716
1
$begingroup$
You can use the result of the user drhab below and proof e.g. by induction.
$endgroup$
– user90369
Mar 28 at 16:53
add a comment |
1
$begingroup$
You can use the result of the user drhab below and proof e.g. by induction.
$endgroup$
– user90369
Mar 28 at 16:53
1
1
$begingroup$
You can use the result of the user drhab below and proof e.g. by induction.
$endgroup$
– user90369
Mar 28 at 16:53
$begingroup$
You can use the result of the user drhab below and proof e.g. by induction.
$endgroup$
– user90369
Mar 28 at 16:53
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In general we have:$$sum_i+j=kbinomirbinomjs=binomk+1r+s+1$$
For a proof of that see here.
Setting $s=0$ we get:$$sum_i=r^kbinomir=binomk+1r+1$$
which get the looks of the summation in your question.
Based on this we find:$$sum_k=0^Nbinomk+n-1n-1=sum_k=n-1^N+n-1binomkn-1=binomN+nn$$
answered Mar 28 at 16:44
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
We use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series. This way we can write for instance
beginalign*
binomnk=[z^k](1+z)^ntag1
endalign*
We obtain
beginalign*
colorbluesum_k=0^Nbinomk+n-1n-1&=sum_k=0^N[z^n-1](1+z)^k+n-1tag2\
&=[z^n-1](1+z)^n-1sum_k=0^N(1+z)^ktag3\
&=[z^n-1](1+z)^n-1frac(1+z)^N+1-1(1+z)-1tag4\
&=[z^n]left((1+z)^N+n-(1+z)^n-1right)tag5\
&,,colorblue=binomN+nntag6
endalign*
Comment:
In (2) we apply the coefficient of operator according to (1).
In (3) we factor out terms which do not depend on $k$.
In (4) we use the finite geometric series formula.
In (5) we collect terms and apply the rule $[z^p-q]A(z)=[z^p]z^qA(z)$.
In (6) we select the coefficient of $z^n$.
answered Mar 28 at 17:00
Markus ScheuerMarkus Scheuer
63.6k460152
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In general we have:$$sum_i+j=kbinomirbinomjs=binomk+1r+s+1$$
For a proof of that see here.
Setting $s=0$ we get:$$sum_i=r^kbinomir=binomk+1r+1$$
which get the looks of the summation in your question.
Based on this we find:$$sum_k=0^Nbinomk+n-1n-1=sum_k=n-1^N+n-1binomkn-1=binomN+nn$$
answered Mar 28 at 16:44
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
In general we have:$$sum_i+j=kbinomirbinomjs=binomk+1r+s+1$$
For a proof of that see here.
Setting $s=0$ we get:$$sum_i=r^kbinomir=binomk+1r+1$$
which get the looks of the summation in your question.
Based on this we find:$$sum_k=0^Nbinomk+n-1n-1=sum_k=n-1^N+n-1binomkn-1=binomN+nn$$
answered Mar 28 at 16:44
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
In general we have:$$sum_i+j=kbinomirbinomjs=binomk+1r+s+1$$
For a proof of that see here.
Setting $s=0$ we get:$$sum_i=r^kbinomir=binomk+1r+1$$
which get the looks of the summation in your question.
Based on this we find:$$sum_k=0^Nbinomk+n-1n-1=sum_k=n-1^N+n-1binomkn-1=binomN+nn$$
answered Mar 28 at 16:44
drhabdrhab
104k545136
$endgroup$
In general we have:$$sum_i+j=kbinomirbinomjs=binomk+1r+s+1$$
For a proof of that see here.
Setting $s=0$ we get:$$sum_i=r^kbinomir=binomk+1r+1$$
which get the looks of the summation in your question.
Based on this we find:$$sum_k=0^Nbinomk+n-1n-1=sum_k=n-1^N+n-1binomkn-1=binomN+nn$$
answered Mar 28 at 16:44
drhabdrhab
104k545136
answered Mar 28 at 16:44
drhabdrhab
104k545136
answered Mar 28 at 16:44
drhabdrhab
104k545136
answered Mar 28 at 16:44
drhabdrhab
104k545136
104k545136
add a comment |
add a comment |
$begingroup$
We use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series. This way we can write for instance
beginalign*
binomnk=[z^k](1+z)^ntag1
endalign*
We obtain
beginalign*
colorbluesum_k=0^Nbinomk+n-1n-1&=sum_k=0^N[z^n-1](1+z)^k+n-1tag2\
&=[z^n-1](1+z)^n-1sum_k=0^N(1+z)^ktag3\
&=[z^n-1](1+z)^n-1frac(1+z)^N+1-1(1+z)-1tag4\
&=[z^n]left((1+z)^N+n-(1+z)^n-1right)tag5\
&,,colorblue=binomN+nntag6
endalign*
Comment:
In (2) we apply the coefficient of operator according to (1).
In (3) we factor out terms which do not depend on $k$.
In (4) we use the finite geometric series formula.
In (5) we collect terms and apply the rule $[z^p-q]A(z)=[z^p]z^qA(z)$.
In (6) we select the coefficient of $z^n$.
answered Mar 28 at 17:00
Markus ScheuerMarkus Scheuer
63.6k460152
$endgroup$
add a comment |
$begingroup$
We use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series. This way we can write for instance
beginalign*
binomnk=[z^k](1+z)^ntag1
endalign*
We obtain
beginalign*
colorbluesum_k=0^Nbinomk+n-1n-1&=sum_k=0^N[z^n-1](1+z)^k+n-1tag2\
&=[z^n-1](1+z)^n-1sum_k=0^N(1+z)^ktag3\
&=[z^n-1](1+z)^n-1frac(1+z)^N+1-1(1+z)-1tag4\
&=[z^n]left((1+z)^N+n-(1+z)^n-1right)tag5\
&,,colorblue=binomN+nntag6
endalign*
Comment:
In (2) we apply the coefficient of operator according to (1).
In (3) we factor out terms which do not depend on $k$.
In (4) we use the finite geometric series formula.
In (5) we collect terms and apply the rule $[z^p-q]A(z)=[z^p]z^qA(z)$.
In (6) we select the coefficient of $z^n$.
answered Mar 28 at 17:00
Markus ScheuerMarkus Scheuer
63.6k460152
$endgroup$
add a comment |
$begingroup$
We use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series. This way we can write for instance
beginalign*
binomnk=[z^k](1+z)^ntag1
endalign*
We obtain
beginalign*
colorbluesum_k=0^Nbinomk+n-1n-1&=sum_k=0^N[z^n-1](1+z)^k+n-1tag2\
&=[z^n-1](1+z)^n-1sum_k=0^N(1+z)^ktag3\
&=[z^n-1](1+z)^n-1frac(1+z)^N+1-1(1+z)-1tag4\
&=[z^n]left((1+z)^N+n-(1+z)^n-1right)tag5\
&,,colorblue=binomN+nntag6
endalign*
Comment:
In (2) we apply the coefficient of operator according to (1).
In (3) we factor out terms which do not depend on $k$.
In (4) we use the finite geometric series formula.
In (5) we collect terms and apply the rule $[z^p-q]A(z)=[z^p]z^qA(z)$.
In (6) we select the coefficient of $z^n$.
answered Mar 28 at 17:00
Markus ScheuerMarkus Scheuer
63.6k460152
$endgroup$
We use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series. This way we can write for instance
beginalign*
binomnk=[z^k](1+z)^ntag1
endalign*
We obtain
beginalign*
colorbluesum_k=0^Nbinomk+n-1n-1&=sum_k=0^N[z^n-1](1+z)^k+n-1tag2\
&=[z^n-1](1+z)^n-1sum_k=0^N(1+z)^ktag3\
&=[z^n-1](1+z)^n-1frac(1+z)^N+1-1(1+z)-1tag4\
&=[z^n]left((1+z)^N+n-(1+z)^n-1right)tag5\
&,,colorblue=binomN+nntag6
endalign*
Comment:
In (2) we apply the coefficient of operator according to (1).
In (3) we factor out terms which do not depend on $k$.
In (4) we use the finite geometric series formula.
In (5) we collect terms and apply the rule $[z^p-q]A(z)=[z^p]z^qA(z)$.
In (6) we select the coefficient of $z^n$.
answered Mar 28 at 17:00
Markus ScheuerMarkus Scheuer
63.6k460152
answered Mar 28 at 17:00
Markus ScheuerMarkus Scheuer
63.6k460152
answered Mar 28 at 17:00
Markus ScheuerMarkus Scheuer
63.6k460152
answered Mar 28 at 17:00
Markus ScheuerMarkus Scheuer
63.6k460152
63.6k460152
add a comment |
add a comment |
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$begingroup$
You can use the result of the user drhab below and proof e.g. by induction.
$endgroup$
– user90369
Mar 28 at 16:53