find the maclaurian series for f(z) = z^3 / (2 - z)^2
and determine the radius of convergence
The Attempt at a Solution
i found the first 8 terms of the series using the maclaurian formula:
z^3/(2-z)^2 = 0 + 0 + 0 + 1/4 * z^3 + 1/4 * z^4 + 3/16 * z^5 + 1/8 * z^6 + 5/64 * z^7
the problem I'm having is comming up wtih a formula for the coefficients so that I can write it out in compact series form.
another way to write it would be:
3/2 * z^3/3! + 6 * z^4/4! + 45/2 * z^5/5! + 90 * z^6/6! * 1575/4 * z^7/7!
so I need a formula for 3/2, 6, 45/2, 90, 1575/4,...
the series starting at n = 3 [ (some formula) * z^(n)/n! ]
could someone please direct me on how to determine a formula for the coefficients.