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Find the formula for the following sum of binomial coefficients: $$ \sum_{m\ge 0} (-1)^m {\binom{n}{m} }^3 .$$ Could you find the formula for $\sum\limits_{m\ge 0} (-1)^m{\binom{n}{m}}^4$?

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What are n and k here? Are they fixed? –  Wonder Apr 18 '12 at 9:31
    
What is {n \choose m} when m is greater than n? 0? –  Wonder Apr 18 '12 at 9:33
    
$n$ is any fixed positive integer. –  hkju Apr 18 '12 at 9:43
    
0% accept rate? Do you know about accepting answers to your questions? and why it's a good thing to do? –  Gerry Myerson Apr 18 '12 at 11:02

2 Answers 2

up vote 5 down vote accepted

Your sum is a special case of Dixon's well-poised sum

$$\begin{align*} \sum_{k=0}^\infty (-1)^k \binom{n}{k}^3&={}_3 F_2\left({{-n,-n,-n}\atop{1,1}}\mid 1\right)\\ &=\frac{\Gamma\left(1-\frac{n}{2}\right)\Gamma \left(\frac{3 n}{2}+1\right)}{n!\Gamma(1-n)\Gamma\left(\frac{n}{2}+1\right)^2}=\frac{\cos\left(\frac{\pi n}{2}\right)\left(\frac{3n}{2}\right)!}{\left(\left(\frac{n}{2}\right)!\right)^3} \end{align*}$$

Generally,

$$\sum_{k=0}^\infty (-1)^k \binom{n}{k}^p={}_p F_{p-1}\left({{-n,\cdots,-n}\atop{1,\cdots,1}}\mid (-1)^{p+1}\right)$$

since $\dbinom{n}{k}=\dfrac{(-1)^k (-n)_k}{k!}$, and $(1)_k=k!$ and thus

$$\sum_{k=0}^\infty (-1)^k \binom{n}{k}^p=\sum_{k=0}^\infty (-1)^k \left(\frac{(-1)^k (-n)_k}{k!}\right)^p=\sum_{k=0}^\infty ((-1)^{p+1})^k \frac{((-n)_k)^p}{((1)_k)^{p-1} k!}$$

and the hypergeometric form is now noticeable.

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(Didn't have Mathematica handy, so I was lucky I remembered Dixon...) –  J. M. Apr 18 '12 at 11:53
    
Is there any proof by using generating function? –  hkju Apr 18 '12 at 13:55
    
I don't know of any... –  J. M. Apr 18 '12 at 13:57
    
How about ∑m=0n(−1)m(nm)p when p is real non-integer between 0 and 2? Are there any results like Dixon's well-poised sum –  user42217 Sep 22 '12 at 2:24

Binomial coefficient sums are mechanical via Gosper's algorithm and extensions.

Wolfram Alpha gives $$\sum_{m=0}^{\infty} (-1)^m \binom{n}{m}^3 = \frac{(1-n)^\overline{n} (1+\frac{n}{2})^\overline{n}}{n! (1-\frac{n}{2})^\overline{n}}$$ (where I've used slightly different notation for the Pochhammer symbol).

Similarly $$\sum_{m=0}^{\infty} (-1)^m \binom{n}{m}^4 = {}_4F_3(-n, -n, -n, -n;1, 1, 1;-1)$$ is hypergeometric-summable.

Checking for $k=5$ and $k=6$ leads to the hypothesis that $$\sum_{m=0}^{\infty} (-1)^m \binom{n}{m}^k = {}_kF_{k-1}(-n, \ldots, -n;1, \ldots, 1;-1^{k-1})$$ but I don't have time now to try to prove this.

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Are the formulas for other powers (I mean, except 3 and 4) similar, and expressed in terms of hypergeometric functions? –  hkju Apr 18 '12 at 10:48
    
@hkju, see edit. –  Peter Taylor Apr 18 '12 at 10:56
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"I don't have time now to try to prove this." - the proof is quite short; see my answer. –  J. M. Apr 18 '12 at 11:44

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