Calculus of a Single Variable, Chapter 9, 9.10, Section 9.10, Problem 18

Recall binomial series that is convergent when |x|lt1 follows:
(1+x)^k=sum_(n=0)^oo _(k(k-1)(k-2)...(k-n+1))/(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- ...
For the given function f(x) =1/(1+x)^4 , we may apply Law of Exponents: 1/x^n = x^(-n) to rewrite it as:
f(x) = (1+x)^(-4)
This now resembles (1+x)^k for binomial series.
By comparing "(1+x)^k " with "(1+x)^(-4) ", we have the corresponding values:
x=x and k = -4 .
Plug-in the values on the formula for binomial series, we get:
(1+x)^(-4)=sum_(n=0)^oo ((-4)(-4-1)(-4-2)...(-4-n+1))/(n!)x^n
= 1 + (-4)x + ((-4)(-4-1))/(2!) x^2 + ((-4)(-4-1)(-4-2))/(3!)x^3 +((-4)(-4-1)(-4-2)(-4-3))/(4!) x^4- ...
= 1 + (-4)x + ((-4)(-5))/(2!) x^2 + ((-4)(-5)(-6))/(3!)x^3 +((-4)(-5)(-6)(-7))/(4!) x^4- ...
= 1 -4x + 20/(2!) x^2 -120/(3!)x^3 +840/(4!)x^4- ...
= 1- 4x +10x^2 -20x^3 +35x^4- ...
Therefore, the Maclaurin series for the function f(x) =1/(1+x)^4 can be expressed as:
1/(1+x)^4 =1- 4x +10x^2 -20x^3 +35x^4- ...