Calculus of a Single Variable, Chapter 9, 9.10, Section 9.10, Problem 22

Recall a binomial series follows:
(1+x)^k=sum_(n=0)^oo _(k(k-1)(k-2)...(k-n+1))/(n!)x^n
or
(1+x)^k = 1 + kx + (k(k-1))/(2!) x^2 + (k(k-1)(k-2))/(3!)x^3 +(k(k-1)(k-2)(k-3))/(4!)x^4+ ...
To evaluate given function f(x) =1/(2+x)^3 , we may apply 2+x = 2(1+x/2) .
The function becomes:
f(x) =1/(2(1+x/2))^3
Apply Law of Exponents: (x*y)^n = x^n*y^n at the denominator side.
1/(2(1+x/2))^3=1/(2^3(1+x/2)^3)
= 1/(8(1+x/2)^3)
Apply Law of Exponents: 1/x^n = x^(-n) .
f(x) = 1/8(1+x/2)^(-3)
Apply the binomial series on (1+x/2)^(-3) . By comparing "(1+x)^k " with "(1+x/2)^(-3) " the corresponding values are:
x=x/2 and k =-3
Then,
(1+x/2)^(-3) =sum_(n=0)^oo _((-3)(-3-1)(-3-2)...(-3-n+1))/(n!)(x/2)^n
=1 + (-3)x/2 + ((-3)(-3-1))/(2!) (x/2)^2 + ((-3)(-3-1)(-3-2))/(3!)(x/2)^3 +((-3)(-3-1)(-3-2)(-3-3))/(4!)(x/2)^4+...
=1 -(3x)/2 + ((-3)(-4))/(2!) (x^2/4) + ((-3)(-4)(-5))/(3!)(x^3/8) +((-3)(-4)(-5)(-6))/(4!)(x^4/16)- ...
=1 -(3x)/2 +12/(2!) (x^2/4) -60/(3!)(x^3/8) +360/(4!)(x^4/16)- ...
=1 -(3x)/2 +(3x^2)/2 -(5x^3)/4 +(15x^4)/16- ...
Applying (1+x/2)^(-3) =1 -(3x)/2 +(3x^2)/2 -(5x^3)/4 +(15x^4)/16- ..., we get:
1/8(1+x/2)^(-3)=1/8*[1 -(3x)/2 +(3x^2)/2 -(5x^3)/4 +(15x^4)/16-...]
=1/8-(3x)/16 +(3x^2)/16 -(5x^3)/32 +(15x^4)/128- ...
Therefore, the Maclaurin series for the function f(x) =1/(2+x)^3 can be expressed as:
1/(2+x)^3=1/8-(3x)/16 +(3x^2)/16 -(5x^3)/32 +(15x^4)/128- ...