Power Series Expansion for Reciprocal of Cube of 1 + x/Proof 2

Theorem

Let $x \in \R$ such that $-1 < x < 1$.

Then:

$\ds \dfrac 1 {\paren {1 + x}^3}$

$=$

$\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\paren {k + 2} \paren {k + 1} } 2 x^k$

$\ds $

$=$

$\ds 1 - 3 x + 6 x^2 - 10 x^3 + 15 x^4 - \cdots$

$\ds \frac 1 {\paren {1 + x} }$

$=$

$\ds \paren {1 + x}^{-3}$

$\ds $

$=$

$\ds \sum_{k \mathop = 0}^\infty \frac {\paren {-3}^{\underline k} } {k!} x^k$

$\ds $

$=$

$\ds \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {\paren {-3} - j} } {k!} x^k$

$\ds $

$=$

$\ds \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {-\paren {j + 3} } } {\ds \prod_{j \mathop = 1}^k j} x^k$

Definition of Factorial and simplifying

$\ds $

$=$

$\ds \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 1}^k \paren {-\paren {j + 2} } } {\ds \prod_{j \mathop = 1}^k j} x^k$

$\ds $

$=$

$\ds \sum_{k \mathop = 0}^\infty \frac {\paren {-1}^k \ds \prod_{j \mathop = 1}^k \paren {j + 2} } {\ds \prod_{j \mathop = 1}^k j} x^k$

$\ds $

$=$

$\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\paren {k + 2} \paren {k + 1} } 2 x^k$

simplification

$\blacksquare$