# Using the Multinomial Theorem to Calculate a Finite Sum raised to an exponent

I know it's a simple question, but I keep getting different general formulas for the coefficients when I am trying to use the multinomial theorem for the following: $$\left(\sum_{k=0}^{M}\frac{(-x^2)^k}{4^kn^{k/2}k!(1+k)!}\right)^n$$ Please help me to calculate it.

Thank you.

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Unforntunately it's not $(1-k)!$ (W|A), but where it come from? Bessel functions? At least Wolfram says that... –  draks ... Jun 25 '12 at 21:43
It comes from the calculation of integral with Bessel J_1 function. I;ve changed the sum to the finite, maybe it will be easier... –  Alex K. Jun 25 '12 at 21:47
Wolfram can give you some examples. Can one of your formulas reproduce that? –  draks ... Jun 25 '12 at 21:53
Thank you. Well, I've got formulas for n=2 and n=3 right. But I need to get a general formula for the coefficients and I am getting it wrong... –  Alex K. Jun 25 '12 at 22:15
Why not posting your solutions? –  draks ... Jun 26 '12 at 16:00


$$\mbox{Then}\,,\quad \bracks{\,\sum_{k = 0}^{M}{(-x^{2})^{k} \over 4^{k}n^{k/2}k!(1 + k)!}}^{n} = \pars{\sum_{k = 0}^{M}a_{k}z^{k}}^{n}\tag{1}$$

\begin{align} &\pars{\sum_{k = 0}^{M}a_{k}z^{k}}^{n}= \sum_{k_{1} = 0}^{M}a_{k_{1}}z^{k_{1}}\sum_{k_{2} = 0}^{M}a_{k_{2}}z^{k_{2}} \cdots\sum_{k_{n} = 0}^{M}a_{k_{n}}z^{k_{n}}\sum_{\ell = 0}^{nM} \delta_{\ell,k_{1}\ +\ k_{2}\ +\ \cdots\ +\ k_{n}} = \sum_{\ell = 0}^{nM}A_{\ell}z^{\ell} \end{align} where

$$\left.A_{\ell} \equiv \sum_{k_{1}, k_{2},\ldots,k_{n} = 0}^{M} a_{k_{1}}a_{k_{2}}\ldots a_{k_{n}}\right\vert_{\sum\limits_{i\ =\ 1}^{n}k_{i}\ =\ \ell}$$
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