Coefficients Party, Part 7

Let $f(x)=(1-2x)^{\frac{3}{2}}$ . The coefficient of $x^6$ in the Maclaurin expansion of $f(x)$ will be: $\dfrac{f^{(6)}(0)}{6!}$ . The rest is simply finding the $6$ th derivative of the function. We find that: $f^{(6)}(x)=\dfrac{315}{(1-2x)^{\frac{9}{2}}}$ . So: $f^{(6)}(0)=315$ , and so the coefficient is: $\dfrac{315}{6!}=\dfrac{7}{16}\equiv\dfrac{A}{B}$ , implying $A+B=\fbox{23}$ .

The terms in the general binomial expansion $\ ( \ 1 \ + \ u \ )^{p/q}$ have the form $\ \left( \begin{array}{c} \frac{p}{q} \\ k \end{array} \right) \ 1^{(p/q) - k} \ u^k \$ [for which we thank Isaac Newton, although he didn't write it this way]. The "fractional" version of a binomial coefficient writes factorials in the usual way, but here they represent "endless products":

$\left( \begin{array}{c} \frac{p}{q} \\ k \end{array} \right) \ = \ \frac{ \frac{p}{q} !}{k! \ ( \ \frac{p}{q} \ - \ k \ )! } \ = \ \frac{ \frac{p}{q} \cdot ( \ \frac{p}{q} - 1 \ ) \cdot \ ( \ \frac{p}{q} - 2 \ ) \cdot \ \ldots}{k! \ \cdot \ ( \ \frac{p}{q} - k \ ) \cdot \ ( \ \frac{p}{q} - k - 1 \ ) \cdot \ \ldots }$

The term in the binomial expansion $\ ( \ 1 \ + \ [ \ -2x \ ] \ )^{3/2}$ we seek is then found from

$\ \left( \begin{array}{c} \frac{3}{2} \\ 6 \end{array} \right) \ ( \ -2x \ )^6 \ = \ \frac{ \frac{3}{2} !}{6! \ ( \ \frac{3}{2} \ - \ 6 \ )! } \ ( -2 )^6 \ x^6 \ = \ \frac{ \frac{3}{2} !}{6! \ ( \ -\frac{9}{2} \ )! } \ 2^6 \ x^6$

$= \ \frac{ \frac{3}{2} \ \cdot \ \frac{1}{2} \ \cdot \ ( \ -\frac{1}{2} \ ) \ \cdot \ ( \ -\frac{3}{2}\ ) \ \cdot \ ( \ -\frac{5}{2} \ ) \ \cdot \ ( \ -\frac{7}{2} \ ) }{6! } \ 2^6 \ x^6 \ = \ \frac{ 3 \ \cdot \ 3 \ \cdot \ 5 \ \cdot \ 7 \ \cdot \ (-1)^4 }{6! \ \cdot \ 2^6} \ 2^6 \ x^6 \ = \ \frac{ 315 }{720} \ x^6 \ = \ \frac{ 7 }{16} \ x^6$ .

The requested sum of numerator and denominator (with the coefficient already in lowest terms) is thus 23.