To find a Maclaurin series for #f(x) = (1)/(2-3x)#, it is probably easier to take a few derivatives in order to see a pattern evolve. Remember that a Maclaurin series is just a special case of a Taylor series centered at #x=0# given by the infinite sum of
#f(0) + (f^(1)(0))/(1!) x + (f^(2)(0))/(2!) x^2 + (f^(3)(0))/(2!) x^3 + ... = sum_(n=0)^(∞) (f^(n)(0)x^n)/(n!)#
As we can see, we'd need a few derivatives to get started.
Derivatives:
#f(x) = (1)/(2-3x) #
#f^(1)(x) = -(2-3x)^(-2)(-3) = (3)/(2-3x)^(2)#
#f^(2)(x) = -6(2-3x)^(-3)(-3) = (18)/(2-3x)^(3) #
#f^(3)(x) = -54(2-3x)^(-4)(-3) = (162)/(2-3x)^(4) #
#f^(4)(x) = -648(2-3x)^(-5)(-3) = (1944)/(2-3x)^(5)#
Evaluating our derivatives at #x=0#:
#f(0) = (1)/(2^1) = 1/2#
# f^(1)(0) = (3)/(2^2) = 3/4#
# f^(2)(0) = (18)/(2^3) = 18/8 #
# f^(3)(0) = (162)/(2^4) = 162/16#
# f^(4)(0) = (1944)/(2^5) = 1944/32#
We can now write out all terms as an infinite sum:
#f(x) = (1/2)/(0!) + (3/4)/(1!) x + (18/8)/(2!) x^2 + (162/16)/(3!) x^3 + (1944/32)/(4!) x^4 + ...#
If we simplify our denominators and numerators we get
#f(x) = 1/2 + 3/4 x + 18/16 x^2 + 162/96 x^3 + 1944/768 x^4 + ...#
And even after more simplification we finally get our infinite sum
#f(x) = 1/2 + 3/4 x + 9/8 x^2 + 27/16 x^3 + 81/32 x^4 + ... = sum_(n=0)^(∞) (3^n x^n)/(2^(n+1))#
