Math Formulas || Series Formulas || Binomial Series

$\begin{aligned} (a + x)^n &= a^n + na^{n-1} + \dfrac{n(n-1)}{2!} a^{n-2}x^2 + \dfrac{n(n-1)(n-2)}{3!}a^{n-3}x^3+\cdots \\ &= a^n + { n \choose 1} a^{n-1}x + { n \choose 2} a^{n-2}x^2 + { n \choose 3} a^{n-3}x^3 + \cdots \end{aligned}$
$\dfrac{1}{1-x} =(1-x)^{-1}= 1 + x + x^2 +x^3 + \cdots \quad |x|<1 $
$\dfrac{1}{1+x} =(1+x)^{-1}= 1 - x + x^2 - x^3 + \cdots \quad|x|<1 $
$\dfrac{1}{(1+x)^2} =(1+x)^{-2}= 1 - 2x + 3x^2 - 4x^3 + \cdots \quad|x|<1$
$\dfrac{1}{(1-x)^2} =(1-x)^{-2}= 1 + 2x + 3x^2 +4x^3 + \cdots \quad |x|<1 $
$(1 + x)^{-1/2} = 1 - \dfrac{1}{2} \ x + \dfrac{1\cdot 3}{2\cdot 4} \ x^2 - \dfrac{1\cdot 3 \cdot 5}{2\cdot 4 \cdot 6} \ x^3 + \cdots \quad | x| \leq 1 $
$(1 + x)^{1/2} = 1 + \dfrac{1}{2} \ x - \dfrac{1}{2\cdot 4} \ x^2 + \dfrac{1\cdot 3}{2\cdot 4 \cdot 6} \ x^3 + \cdots \quad |x| \leq 1$
$(1 - x)^{1/2} = 1 - \dfrac{1}{2} \ x - \dfrac{1\cdot 3}{2\cdot 4} \ x^2 - \dfrac{1\cdot 3 \cdot 5}{2\cdot 4 \cdot 6} \ x^3 - \cdots \quad | x| \leq 1$
$(1 - x)^{-1/2} = 1 + \dfrac{1}{2} \ x + \dfrac{1\cdot 3}{2\cdot 4} \ x^2 + \dfrac{1\cdot 3 \cdot 5}{2\cdot 4 \cdot 6} \ x^3 + \cdots \quad | x| \leq 1 $