## Calculus 8th Edition

The series is convergent if $\frac{19}{4} \lt x \lt \frac{21}{4}$ and the sum is $\frac{1}{4x-19}$.
Given: $\Sigma^{\infty}_{n=0} (-4)^{n}(x-5)^{n}$ $\Sigma^{\infty}_{n=0} (-4)^{n}(x-5)^{n} = \Sigma^{\infty}_{n=0}(-4x+20)^{n}$ Here, $a=1$ and $r=(-4x+20)^{n}$ Also, $|r| \lt 1$ $|-4x + 20| \lt 1$ $-1 \lt -4x +20 \lt 1$ $-21 \lt -4x \lt -19$ $\frac{19}{4} \lt x \lt \frac{21}{4}$ The series is convergent if $\frac{19}{4} \lt x \lt \frac{21}{4}$ The sum can be calculated as follows: $\Sigma^{\infty}_{n=0} (-4)^{n} (x-5)^{n} = \frac{a}{1-r}$ $=\frac{1}{1-(-4x+20)}$ $=\frac{1}{4x-19}$