Finding the power series of: $G(t) = \frac{1+3t}{5-t^2}$

Question

How would I go about finding the power series of the function;

$G(t) = \frac{1+3t}{5-t^2}$

Answer 1

Write it as $G(t)=\frac{1}{5-t^2}+\frac{3t}{5-t^2}$. For the first fraction, this can be rewritten as $$ \frac{1}{5-t^2}=\frac{1}{5}\cdot\frac{1}{1-\tfrac{t^2}{5}} $$ which is a simple geometric series which can be expanded. Doing the same thing for the second sum, we get $$ \frac{3t}{5-t^2}=\frac{3t}{5}\cdot\frac{1}{1-\tfrac{t^2}{5}} $$ Now just expand the geometric series, multiply through, and add the two sums.

Answer 2

$$\frac { 1+3t }{ 5-t^{ 2 } } =\frac { A }{ 5-t } +\frac { B }{ 5+t } \\ 1+3t=A\left( 5+t \right) +B\left( 5-t \right) =\left( A-B \right) t+5A+5B\\ \\ \begin{cases} A-B=3 \\ A+B=0.2 \end{cases}\Rightarrow \begin{cases} A=1.6 \\ B=-1.4 \end{cases}$$ for $\left| t \right| <1$ we have $$\\ \frac { 1+3t }{ 5-t^{ 2 } } =\frac { 1.6 }{ 5-t } -\frac { 1.4 }{ 5+t } =\frac { 1.6 }{ 5 } \frac { 1 }{ 1-\left( \frac { t }{ 5 } \right) } -\frac { 1.4 }{ 5 } \frac { 1 }{ 1-\left( -\frac { t }{ 5 } \right) } =\frac { 1.6 }{ 5 } \sum _{ n=0 }^{ \infty }{ { \left( \frac { t }{ 5 } \right) }^{ n } } -\frac { 1.4 }{ 5 } \sum _{ n=0 }^{ \infty }{ { \left( -1 \right) }^{ n } } { \left( \frac { t }{ 5 } \right) }^{ n }=\\ =\sum _{ n=0 }^{ \infty }{ { \left( \frac { t }{ 5 } \right) }^{ n } } \left( \frac { 1.6-1.4{ \left( -1 \right) }^{ n } }{ 5 } \right) $$