Explain how to use the geometric series $g ( x ) = \frac { 1 } { 1 - x } = \sum _ { n = 0 } ^ { \infty } x ^ { n } , | x |

Question

Explain how to use the geometric series $g ( x ) = \frac { 1 } { 1 - x } = \sum _ { n = 0 } ^ { \infty } x ^ { n } , | x | < 1 \text { to find the }$ series for the function $\frac { 9 } { 1 + x }$.
A) replace $x$ with $\frac { 9 } { ( - x ) }$
B) replace $x$ with $( - x )$ and multiply the series by 9
C) replace $x$ with $\frac { 1 } { x }$ and divide the series by 9
D) replace $x$ with $( - x )$ and divide the series by 9
E) replace $x$ with $\frac { 9 } { x }$

Quizplus · Accepted Answer

We can use the formula for a geometric series, $g(x) = \frac{1}{1-x} = \sum_{n=0}^{\infty}x^n$, to find the series for $\frac{9}{1+x}$ by first expressing the denominator in the form of $1-x$. We have:

$$\frac{9}{1+x} = 9\cdot\frac{1}{1+(-x)}$$

We can now replace $x$ by $-x$ and use the geometric series formula:

\begin{align*}
9\cdot\frac{1}{1+(-x)} &= 9\cdot \sum_{n=0}^\infty (-x)^n\
&=9\cdot \sum_{n=0}^\infty (-1)^n x^n
\end{align*}

Multiplying the series by 9 gives us the series for $\frac{9}{1+x}$:

$$\frac{9}{1+x} = \sum_{n=0}^\infty 9(-1)^n x^n$$

Explain How to Use the Geometric Series g(x)=11−x=∑n=0∞xn,∣x∣<1 to find the g ( x ) = \frac { 1 } { 1 - x } = \sum _ { n = 0 } ^ { \infty } x ^ { n } , | x | < 1 \text { to find the }g(x)=1−x1​=∑n=0∞​xn,∣x∣<1 to find the

Explain How to Use the Geometric Series $g ( x ) = \frac { 1 } { 1 - x } = \sum _ { n = 0 } ^ { \infty } x ^ { n } , | x | < 1 \text { to find the }$