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I have run across the following multinomial series: $$ \sum_{j = a}^{N} \binom{N}{j} \binom{j}{a} d^{-j} $$ Here, $d>1$. This seems like a formula which has either a well-known identity, or which has no further closed form or simplification. Can anyone shed any light on this formula, or possibly point me to a standard reference?

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for $a=0$ we have $(1+1/d)^N$ (not that this helps)... –  yoyo Mar 3 '11 at 20:31
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4 Answers

up vote 7 down vote accepted

$$\sum_{j = a}^{N} \binom{N}{j} \binom{j}{a} d^{-j} = \sum_{j = a}^{N} \binom{N}{a} \binom{N-a}{j-a} d^{-j} = \binom{N}{a} \sum_{j = 0}^{N-a} \binom{N-a}{j} d^{-j-a} $$ $$= d^{-a} \binom{N}{a} \sum_{j = 0}^{N-a} \binom{N-a}{j} d^{-j} = d^{-a} \binom{N}{a} (1+d^{-1})^{N-a}.$$ The first step uses what's called trinomial revision. See, for example, Concrete Mathematics, p. 174. The last step uses the binomial theorem.

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You can use $${N \choose j}{j \choose a} = \frac{ N!}{j! (N-j)!} \frac{j!}{a! (j-a)!} =\frac{N!}{a!(N-a)!} \frac{(N-a)!}{(j-a)! (N-j)!} = {N \choose a}{ N-a \choose j-a}$$ to write your expression as $${N \choose a} \sum_{j=a}^N { N-a \choose j-a} d^{-j}= {N \choose a} d^{-a} \sum_{j=0}^{N-a} { N -a \choose j} d^{-j} = {N \choose a} d^{-a} (1+1/d)^{N-a}.$$ (using binomial series)

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Since you mentioned the word "multinomial", here is another way, using the Multinomial theorem.

Expand

$$(1 + x/d + x)^n = \sum_{p+q+r=n} {n \choose p,q,r} 1^p \cdot (x/d)^q \cdot x^r$$

$$= \sum_{a=0}^{n} \sum_{j=a}^{n} {n \choose a,j-a,n-j} 1^{a} \cdot (x/d)^{j-a} \cdot x^{n-j}$$

$$ = \sum_{a=0}^{n} \left(\sum_{j=a}^{n} {n \choose j} {j \choose a} d^{a-j}\right) x^{n-a}$$

Expand $\displaystyle (1 + x(1+1/d))^n$ using binomial theorem and equate the coefficient of $\displaystyle x^{n-a}$ on both sides to get

$$ {n \choose a} (1 + 1/d)^{n-a} = \sum_{j=a}^{n} {n \choose j} {j \choose a} d^{a-j}$$

and so

$$ {n \choose a} d^{-a} (1 + 1/d)^{n-a} = \sum_{j=a}^{n} {n \choose j} {j \choose a} d^{-j}$$

Here are a couple more answers which use similar ideas:

Beautiful identity: $\sum_{k=m}^n (-1)^{k-m} \binom{k}{m} \binom{n}{k} = \delta_{mn}$

Non-probabilistic proofs of a binomial coefficient identity from a probability question

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Answer: $$\binom{N}{a} d^{-a}(1+d^{-1})^{N-a} \; .$$

I'll leave it up to you to check it.

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