Summation with binomial coefficient
by fungarwai, Jan 12, 2019, 2:04 PM
Summation with one binomial coefficient
from Binomial theorem
which is Hockey-stick identity
Summation with two binomial coefficient
which is Vandermonde's identity
Summation with three binomial coefficient
which is the Li Shanlan identity
Summation with multiple binomial coefficient
where
It is the Generalized Vandermonde's identity
Named as Vandermonde type convolution formula in
Combinatorial identities associated with new families of the numbers and polynomials and their approximation values
(Theorem 5.4 of Section 5)
The probability of the sum of negative binomial random variables also implies that
Reference: Negative binomial distribution - sum of two random variables
It can be further generalised as:
![$\displaystyle \sum_{\sum c_in_i=n} \binom{n_1}{m_1}\binom{n_2}{m_2}\cdots
\binom{n_k}{m_k}=[x^{n-\sum c_im_i}]\prod_{i=1}^k\frac{1}{(1-x^{c_i})^{m_i+1}}$](//latex.artofproblemsolving.com/0/f/3/0f3f5f0cccbed5ee444ad5c081522482ac03a41a.png)
from Binomial theorem
which is Hockey-stick identity
Summation with two binomial coefficient
which is Vandermonde's identity
Summation with three binomial coefficient
which is the Li Shanlan identity
Summation with multiple binomial coefficient
where
It is the Generalized Vandermonde's identity
to get the result
Named as Vandermonde type convolution formula in
Combinatorial identities associated with new families of the numbers and polynomials and their approximation values
(Theorem 5.4 of Section 5)
The probability of the sum of negative binomial random variables also implies that
Reference: Negative binomial distribution - sum of two random variables
for 
solutions for 
,
is 
It can be further generalised as:
![$\displaystyle \sum_{\sum c_in_i=n} \binom{n_1}{m_1}\binom{n_2}{m_2}\cdots
\binom{n_k}{m_k}=[x^{n-\sum c_im_i}]\prod_{i=1}^k\frac{1}{(1-x^{c_i})^{m_i+1}}$](http://latex.artofproblemsolving.com/0/f/3/0f3f5f0cccbed5ee444ad5c081522482ac03a41a.png)
![$\displaystyle \sum_{c_1n_1+c_2n_2+\cdots+c_kn_k=n} \binom{n_1+m_1}{m_1}\binom{n_2+m_2}{m_2}\cdots
\binom{n_k+m_k}{m_k}=[x^n]\prod_{i=1}^k\frac{1}{(1-x^{c_i})^{m_i+1}}$](http://latex.artofproblemsolving.com/9/7/2/9725c9d81484580eb57d9b5af074e5a4915a30ce.png)
![$\displaystyle \sum_{c_1n_1+c_2n_2+\cdots+c_kn_k=n} \binom{n_1}{m_1}\binom{n_2}{m_2}\cdots
\binom{n_k}{m_k}=[x^{n-\sum_{i=1}^k c_im_i}]\prod_{i=1}^k\frac{1}{(1-x^{c_i})^{m_i+1}}$](http://latex.artofproblemsolving.com/7/4/5/74586763bae2fc73fb3416c03c734cd3f39c4b08.png)
,
are triples of positive integers.
^2(1+x^2+x^4)^2(1+x+x^2+x^3+x^4+x^5)^2\sum_{n=0}^\infty \binom{n+5}{5}x^{6n}$](http://latex.artofproblemsolving.com/f/6/e/f6eb113e7b8c82361610ca9f4f4ec3d23754da24.png)
^2(1+x^2+x^4)^2(1+x+x^2+x^3+x^4+x^5)^2\sum_{k=0}^\infty \binom{k+5}{5}x^{6k}$](http://latex.artofproblemsolving.com/4/c/a/4cab818db089ae4528d2a491360c69c5ae281818.png)
This post has been edited 27 times. Last edited by fungarwai, Apr 27, 2025, 7:24 AM
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