# How do you find taylor polynomials of degree n approximating 5/(2-2x) for x near 0?

Jul 28, 2016

$\frac{5}{2 - 2 x} = {\sum}_{n = 0}^{\infty} \frac{5}{2} {x}^{n}$

#### Explanation:

If $\left\mid x \right\mid < 1$ then:

$\frac{1}{1 - x} = {\sum}_{n = 0}^{\infty} {x}^{n}$

Hence:

$\frac{5}{2 - 2 x} = \frac{5}{2 \left(1 - x\right)} = \frac{5}{2} \cdot \frac{1}{1 - x} = \frac{5}{2} {\sum}_{n = 0}^{\infty} {x}^{n} = {\sum}_{n = 0}^{\infty} \frac{5}{2} {x}^{n}$