$\text { Use the power series } \frac { 1 } { 1 + x } = \sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n } x ^ { n } \text { to determine a power series centered at } 0$
for the function $f ( x ) = - \frac { 8 } { ( 8 x + 1 ) ^ { 2 } } = \frac { d } { d x } \left[ \frac { 1 } { 8 x + 1 } \right]$
A) $\sum _ { n = 0 } ^ { \infty } ( - 8 ) ^ { n } n x ^ { n - 1 }$
B)$\sum _ { n = 0 } ^ { \infty } ( 8 ) ^ { n } n x ^ { n - 1 }$
C)$\sum _ { n = 0 } ^ { \infty } ( - 8 ) ^ { n } x ^ { n }$
D)$\sum _ { n = 0 } ^ { \infty } ( - 8 ) ^ { n } x ^ { n }$.
E)$\sum _ { n = 0 } ^ { \infty } ( - 8 ) ^ { n } ( n - 1 ) x ^ { n - 1 }$.

Question

$\text { Use the power series } \frac { 1 } { 1 + x } = \sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n } x ^ { n } \text { to determine a power series centered at } 0$ 
for the function $f ( x ) = - \frac { 8 } { ( 8 x + 1 ) ^ { 2 } } = \frac { d } { d x } \left[ \frac { 1 } { 8 x + 1 } \right]$
A) $\sum _ { n = 0 } ^ { \infty } ( - 8 ) ^ { n } n x ^ { n - 1 }$
B)$\sum _ { n = 0 } ^ { \infty } ( 8 ) ^ { n } n x ^ { n - 1 }$
C)$\sum _ { n = 0 } ^ { \infty } ( - 8 ) ^ { n } x ^ { n }$
D)$\sum _ { n = 0 } ^ { \infty } ( - 8 ) ^ { n } x ^ { n }$.
E)$\sum _ { n = 0 } ^ { \infty } ( - 8 ) ^ { n } ( n - 1 ) x ^ { n - 1 }$.

Quizplus · Accepted Answer

The given function $f(x) = -\frac{8}{(8x + 1)^2}$ can be derived from the power series of $\frac{1}{1+x}$ by substituting $x$ with $8x$ and differentiating. The differentiation increases the power of $n$ by multiplying it, leading to the correct series representation.

For the Function f(x)=−8(8x+1)2=ddx[18x+1]f ( x ) = - \frac { 8 } { ( 8 x + 1 ) ^ { 2 } } = \frac { d } { d x } \left[ \frac { 1 } { 8 x + 1 } \right]f(x)=−(8x+1)28​=dxd​[8x+11​]

For the Function $f ( x ) = - \frac { 8 } { ( 8 x + 1 ) ^ { 2 } } = \frac { d } { d x } \left[ \frac { 1 } { 8 x + 1 } \right]$