What kind of Binomial is this?

What kind of Binomial? Well that's a binomial with a fractional (and negative) index...

$(1-x)^{-n} = \displaystyle \sum_{k=0}^{\infty} \dbinom{n+r-1}{r} x^r$

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$\Bigl( 1 - \dfrac{1}{2} \Bigr) ^\frac{-2}{3} = 1+ \Bigl(\frac{-2}{3}\Bigr)\Bigl(\dfrac{-1}{2}\Bigr) + \dfrac{\frac{-2}{3} \frac{-5}{3} }{2!} \Bigl(\dfrac{-1}{2} \Bigr) ^2 + \dfrac{\frac{-2}{3} \frac{-5}{3} \frac{-8}{3}}{3!} \Bigl(\dfrac{-1}{2}\Bigr)^3 +....$

That's what is asked in the problem, but in the form

$\quad = 1 + \dfrac{2}{3} \Bigl(\frac{1}{2} \Bigr)^1+ \dfrac{2\times 5}{3\times 6} \Bigl(\frac{1}{2}\Bigr)^2+\dfrac{2\times 5\times 8}{3\times 6\times 9} \Bigl(\frac{1}{2}\Bigr)^3+....$

which is just what is asked, so answer is $\Bigl(1-\frac{1}{2}\Bigr)^{\frac{-2}{3}} = 2^{\dfrac{2}{3}} = 4^{\dfrac{1}{3}}$

Answer $4+3=\boxed{7}$

Recall that for $|x| <1$
$\begin{aligned} & (1+ x)^{\frac{1}{n}} = 1 + nx + \frac{n(n-1)x^2}{2!} + \frac{n(n-1)(n-2)x^3}{3!} + \cdots \\& (1+x)^{\frac{1}{n}} = 1 + \frac{2}{3}\left(\frac{1}{2}\right) + \frac{2\times 5}{3\times 6}\left(\frac{1}{2}\right)^2 + \frac{2\times 5\times 8}{3\times 6\times 9 }\left(\frac{1}{2}\right)^3 \end{aligned}$

Now on comparing we note that $\begin{aligned}& nx = \frac{2}{3}\left(\frac{1}{2}\right) = \frac{1}{3}\\& \frac{n(n-1)x^2}{2!}= \frac{2\times 5}{3\times 6}\left(\frac{1}{2}\right)^2 = \frac{5}{3\times 6\times 2} \\& \frac{(nx)(nx -x)}{2} = \frac{5}{18\times 2 } \\& nx-x = \frac{5}{6} \qquad x = - \frac{5}{6} + \frac{1}{3} = - \frac{1}{2} \end{aligned}$ plugging the value of $x$ in first equation we get $nx = \frac{1}{3} \qquad n = -\frac{ 2}{3}$

Hence the infinite series can be expressed as $\begin{aligned} & (1 +x)^{\frac{1}{n}} = 1 + nx + \frac{n(n-1)x^2}{2!} + \frac{n(n-1)(n-2)x^3}{3!} + \cdots \\& (1-\frac{1}{2})^{-\frac{2}{3}} = 1 + \frac{2}{3}\left(\frac{1}{2}\right) + \frac{2\times 5}{3\times 6}\left(\frac{1}{2}\right)^2 + \frac{2\times 5\times 8}{3\times 6\times 9 }\left(\frac{1}{2}\right)^3 \\& 4^{\frac{1}{3}} = 1 + \frac{2}{3}\left(\frac{1}{2}\right) + \frac{2\times 3}{3\times 6}\left(\frac{1}{2}\right)^2 + \frac{2\times 3 \times 8}{3\times 6\times 9}\left(\frac{1}{2}\right)^3 \cdots + \end{aligned}$

Since $4$ and $3$ are co-prime integers So $a+b = \boxed{7}$ .