Closed form for a sum

Question

Please, i need help with this example, step by step.

Calculate the value of the next summation, i.e. express simple formula without the sum:

$$\sum_{n_1 + n_2 + n_3 + n_4 = 5} \frac{6^{n_2-n_4} (-7)^{n_1}}{n_1! n_2! n_3! n_4!}$$

I think the formula is

$\sum_{n1 + n2 + n3 + n4 = n} = \frac{n!* x^{n_1} * x2^{n_2} * x3^{n_3}* x4^{n_4}}{n2! n2! n3! n4!}$

I don't know how to proceed, i stuck here.

Here is prtscr: http://prntscr.com/l8gluf

Answer 1

We consider the multinomial theorem in the form \begin{align*} \left(x_1+x_2+x_3+x_4\right)^5&=\sum_{{n_1+n_2+n_3+n_4=5}\atop{n_j\geq 0, 1\leq j\leq 4}} \binom{5}{n_1,n_2,n_3,n_4}x_1^{n_1}x_2^{n_2}x_3^{n_3}x_4^{n_4}\\ &=\sum_{{n_1+n_2+n_3+n_4=5}\atop{n_j\geq 0, 1\leq j\leq 4}}\frac{5!}{n_1!n_2!n_3!n_4!}x_1^{n_1}x_2^{n_2}x_3^{n_3}x_4^{n_4}\tag{1} \end{align*}

We obtain applying (1) \begin{align*} \color{blue}{\sum_{{n_1+n_2+n_3+n_4=5}\atop{n_j\geq 0, 1\leq j\leq 4}}}& \color{blue}{\frac{6^{n_2-n_4} (-7)^{n_1}}{n_1! n_2! n_3! n_4!}}\\ &=\frac{1}{5!}\sum_{{n_1+n_2+n_3+n_4=5}\atop{n_j\geq 0, 1\leq j\leq 4}}\frac{5!}{n_1! n_2! n_3! n_4!} (-7)^{n_1}6^{n_2}1^{n_3}\left(\frac{1}{6}\right)^{n_4} \\ &=\frac{1}{5!}\left(-7+6+1+\frac{1}{6}\right)^5\\ &\,\,\color{blue}{=\frac{1}{933\,120}} \end{align*}

Answer 2

Hint. The multinomial theorem is

$$(x_1 + x_2 + \ldots + x_k)^n = \sum_{n_1 + \cdots + n_k = n} {n \choose n_1, \ldots, n_k} x_1^{n_1} x_2^{n_2} \cdots x_k^{n_k}$$

Now let $k=4$, $n=5$ and find values of $x_1,x_2,x_3,x_4$ to make the right-hand side summand look a bit like the summand in your expression. Note that $${n \choose n_1, \ldots, n_k} = \frac{n!}{n_1! \cdots n_k!}$$ so you get the $n_1!n_2!n_3!n_4!$ in the denominator for "free"...

Answer 3

Can i do this with expression?

$$\sum_{n_1 + n_2 + n_3 + n_4 = 5} \frac{5! *(-7)^{n_1} * 6^{n_2} * 1^{n3} * (1/6)^{n_4}}{n_1! n_2! n_3! n_4!}$$

What next? I just put value into original expression