Calculus of a Single Variable, Chapter 9, 9.10, Section 9.10, Problem 17

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!)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- ...
For given function f(x) =1/(1+x)^2 , we may apply Law of Exponents: 1/x^n = x^(-n) to rewrite it as:
f(x) = (1+x)^(-2)
This now resembles (1+x)^k for binomial series.
By comparing "(1+x)^k " with "(1+x)^(-2) ", we have the corresponding values:
x=x and k = -2 .
Plug-in the values on the formula for binomial series, we get:
(1+x)^(-2)=sum_(n=0)^oo (-2(-2-1)(-2-2)...(-2-n+1))/(n!)x^n
= 1 + (-2)x + (-2(-2-1))/(2!) x^2 + (-2(-2-1)(-2-2))/(3!)x^3 +(-2(-2-1)(-2-2)(-2-3))/(4!) x^4- ...
= 1 -2x + 6/(2!) x^2 -24/(3!)x^3 +120/(4!)x^4- ...
= 1- 2x +3x^2 -4x^3 +5x^4- ...
or sum_(n=0)^oo (-1)^n (n+1)x^n
Therefore, the Maclaurin series for the function f(x) =1/(1+x)^2 can be expressed as:
1/(1+x)^2 =sum_(n=0)^oo (-1)^n (n+1)x^n
or
1/(1+x)^2 =1- 2x +3x^2 -4x^3 +5x^4- ...