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3 votes
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$(1-x)^{n+a} \sum_{j=0}^\infty \binom{n+j-1}{j}\binom{n+j}{a} x^j = \sum_{j=0}^a \binom{n}{a-j}\binom{a-1}{j} x^j$

With standard notations (Pochhammer's symbol, Gaussian Hypergeometric Function, $\Gamma$-function and Roy's strategy in https://dx.doi.org/10.2307/2323500 ) the left hand side is essentially $$ \sum_{...
R. J. Mathar's user avatar
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3 votes
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Spivak Exercise, Prove Vandermonde's Identity $\sum_{k=0}^{l}\binom{n}{k}\binom{m}{l-k}=\binom{n+m}{l}$

No, you should write $$\sum_{k=0}^n\binom nk x^k \cdot \sum_{j=0}^m \binom mj x^j = \sum_{s=0}^{m+n} \binom{m+n}s x^s.$$ Now compare the coefficients of $x^l$ on both sides. So you will need $j+k=l$, ...
Ted Shifrin's user avatar
3 votes

Deriving equation (3-66) of Papoulis and Pillai

We have that $$P_{2n} = \sum_{k=0}^{n-1} {2n\choose k} p^{2n-k} q^k = p^{2n} [z^{n-1}] \frac{1}{1-z} (1+qz/p)^{2n} \\ = p^2 [z^{n-1}] \frac{1}{1-p^2z} (1+pqz)^{2n} = p^2 [z^{n+1}] z^2 \frac{1}{1-p^2z} ...
Marko Riedel's user avatar
2 votes

Is there a general formula for $F(n,k) = \sum_{i=0}^{n}\binom{n}{i}^k$?

For $n=3$, they are the Franel numbers (sequence $A000172$ in $OEIS$). In fact, there is a general formula which involves generalized hypergeometric functions (which are summations up to infinity !!). ...
Claude Leibovici's user avatar
2 votes

Deriving equation (3-66) of Papoulis and Pillai

Here is my development of the argument given in Papoulis and Pillai. First,we may use the binomial expansion to write $$(p+q)^{2n}=\sum_{k=0}^{2n}\binom{2n}{k}p^k q^{2n-k}=\sum_{k=0}^{n}\binom{2n}{k}p^...
Semiclassical's user avatar
2 votes

The upper bound of $\sum_{k=0}^{m} \binom{n}{k}x^k(1-x)^{n-k}$

Here is an approximation approach. Instead of refering to a beta distribution, consider a binomial $Bin(n,p)$ random variable $X$. In this framework : $$\sum_{k=0}^{m} \binom{n}{k}x^k(1-x)^{n-k}$$ can ...
Jean Marie's user avatar
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1 vote

$(1-x)^{n+a} \sum_{j=0}^\infty \binom{n+j-1}{j}\binom{n+j}{a} x^j = \sum_{j=0}^a \binom{n}{a-j}\binom{a-1}{j} x^j$

We need to show $$ (1-x)^{n+a} \sum_{m=0}^\infty \binom{n+m-1}{m}\binom{n+m}{a} x^m = \sum_{j=0}^a \binom{n}{a-j}\binom{a-1}{j} x^j. $$ Instead, let's move $(1-x)^{n+a}$ to the RHS using a standard ...
Oleksandr  Kulkov's user avatar
1 vote

$(1-x)^{n+a} \sum_{j=0}^\infty \binom{n+j-1}{j}\binom{n+j}{a} x^j = \sum_{j=0}^a \binom{n}{a-j}\binom{a-1}{j} x^j$

We seek to show that with $n$ and $a$ natural numbers, $$(1-x)^{n+a} \sum_{j\ge 0} {n+j-1\choose j} {n+j\choose a} x^j = \sum_{j=0}^a {n\choose a-j} {a-1\choose j} x^j.$$ Note that we may assume that $...
Marko Riedel's user avatar
1 vote

Deriving equation (3-66) of Papoulis and Pillai

We have the sequence $$ P_{2n} = \sum\limits_{k=n+1}^{2n} \binom{2n}{k} p^k q^{2n-k}. $$ It contains the terms of $(p+q)^{2n}$ for powers of $q$ from $0$ to $n-1$. One concise way to formulate it is $$...
Oleksandr  Kulkov's user avatar
1 vote

Prove $\sum_{k=0}^{r}\binom{n}{k}\binom{n}{r-k}\left(-1\right)^{k}=\left(-1\right)^{\frac{r}{2}}\binom{n}{\frac{r}{2}}$ for $r$ even.

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Felix Marin's user avatar
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