$\text { Find a power series for the function } \frac { 24 x - 35 } { 6 x ^ { 2 } + 29 x - 5 } \text { centered at } 0 \text {. }$
A)
$\sum _ { n = 0 } ^ { \infty } \left( - \frac { 1 } { 5 ^ { n } } + 6 ^ { n + 1 } \right) x ^ { n }$
B)
$\sum _ { n = 0 } ^ { \infty } \left( \frac { x } { 5 ^ { n } } + 6 ^ { n + 1 } \right) x ^ { n }$
C)
$\sum _ { n = 0 } ^ { \infty } \left( \frac { ( - 1 ) ^ { n } } { 5 ^ { n } } + 6 ^ { n + 1 } \right) x ^ { n }$
D)
$\sum _ { n = 0 } ^ { \infty } \left( - \frac { x } { 6 ^ { n } } + 5 ^ { n + 1 } \right) x ^ { n }$
E)
$\sum _ { n = 0 } ^ { \infty } \left( \frac { ( - 1 ) ^ { n } } { 6 ^ { n } } + 5 ^ { n + 1 } \right) x ^ { n }$

Question

$\text { Find a power series for the function } \frac { 24 x - 35 } { 6 x ^ { 2 } + 29 x - 5 } \text { centered at } 0 \text {. }$ 
A) 
$\sum _ { n = 0 } ^ { \infty } \left( - \frac { 1 } { 5 ^ { n } } + 6 ^ { n + 1 } \right) x ^ { n }$
B)
$\sum _ { n = 0 } ^ { \infty } \left( \frac { x } { 5 ^ { n } } + 6 ^ { n + 1 } \right) x ^ { n }$
C)
$\sum _ { n = 0 } ^ { \infty } \left( \frac { ( - 1 ) ^ { n } } { 5 ^ { n } } + 6 ^ { n + 1 } \right) x ^ { n }$
D)
$\sum _ { n = 0 } ^ { \infty } \left( - \frac { x } { 6 ^ { n } } + 5 ^ { n + 1 } \right) x ^ { n }$
E)
$\sum _ { n = 0 } ^ { \infty } \left( \frac { ( - 1 ) ^ { n } } { 6 ^ { n } } + 5 ^ { n + 1 } \right) x ^ { n }$

Quizplus · Accepted Answer

We can use partial fraction decomposition to write the function as $\frac{24x-35}{(2x-1)(3x+5)}=\frac{A}{2x-1}+\frac{B}{3x+5}$ for some constants $A$ and $B$. Solving for $A$ and $B$, we get $A=\frac{3}{7}$ and $B=-\frac{2}{7}$. Therefore, we have
$$\frac{24x-35}{(2x-1)(3x+5)}=\frac{3}{7}\cdot\frac{1}{2x-1}-\frac{2}{7}\cdot\frac{1}{3x+5}$$
Now, we can use geometric series to expand each term:
$$\frac{3}{7}\cdot\frac{1}{2x-1}=\frac{3}{7}\cdot\frac{1}{1-\frac{1-2x}{2x-1}}=\sum_{n=0}^\infty \frac{3(1-2x)^n}{7\cdot 2^{n+1}}x^n$$
$$-\frac{2}{7}\cdot\frac{1}{3x+5}=-\frac{2}{7}\cdot\frac{1}{5}\cdot \frac{1}{1+\frac{3x}{5}}=-\frac{2}{35}\sum_{n=0}^\infty \left(-\frac{3x}{5}ight)^n x^n=\sum_{n=0}^\infty \frac{2(-3)^n}{35\cdot 5^{n+1}}x^n$$
Adding these together, we get
$$\frac{24x-35}{(2x-1)(3x+5)}=\sum_{n=0}^\infty\left(\frac{3(1-2x)^n}{7\cdot 2^{n+1}}-\frac{2\cdot 3^n}{35\cdot 5^{n+1}}ight)x^n=\sum_{n=0}^\infty\left(\frac{(-1)^n}{5^n}+\frac{6^n}{7}ight)x^{n+1}$$
Note that we started the sum from $n=0$ instead of $n=1$ because the original function has a removable discontinuity at $x=\frac{1}{2}$, which means the function is actually analytic (i.e. has a power series representation) on a slightly larger interval that includes $0$. Therefore, the correct choice is $\boxed{	extbf{(C)}}$.

Find a power series for the function 24x−356x2+29x−5 centered at 0. \text { Find a power series for the function } \frac { 24 x - 35 } { 6 x ^ { 2 } + 29 x - 5 } \text { centered at } 0 \text {. } Find a power series for the function 6x2+29x−524x−35​ centered at 0.

$\text { Find a power series for the function } \frac { 24 x - 35 } { 6 x ^ { 2 } + 29 x - 5 } \text { centered at } 0 \text {. }$