$f(x) = \sqrt{x}$
When you do expansion, you get the following:
$$4 + \frac{x-16}{2^3}\times 1! + \frac{(x-16)^2}{2^8}\times 2! + \frac{3(x-16)^3}{2^{13}}\times 3!$$$$+ \frac{15(x-16)^4}{2^{18}}\times 4! + \frac{105(x-16)^5}{2^{23}}\times 5!+\cdots$$
I know the denominator of the summation with E n=1 -> infinity is 2^(5n-2) * n!.
But for the numerator it starts with $c_1 = 0$, $c_2 = 0$, $c_3 = 3$, $c_4 = 15$, $c_5 = 105$ as the coefficients on the $(x-16)^n$. What would be the coefficient in the summation for $(x-16)^n$, as in, what would $(x-16)^n$ multiply by in the numerator of the summation?
