Given the geometric series: $$ 8 x^{2}+4 x^{3}+2 x^{4}+\ldots $$ 3.1 Determine the $$ \mathrm{n}^{ ext {th }} $$ term of the series. 3.2 For what value(s) of $$ x $$ will the series converge? 3.3 Calculate the sum of the series to infinity if $$ x=\frac{3}{2} $$.

Question

Given the geometric series: $$ 8 x^{2}+4 x^{3}+2 x^{4}+\ldots $$ 3.1 Determine the $$ \mathrm{n}^{	ext {th }} $$ term of the series. 3.2 For what value(s) of $$ x $$ will the series converge? 3.3 Calculate the sum of the series to infinity if $$ x=\frac{3}{2} $$.

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To solve the problem, let's break it down step by step.

### 3.1 Determine the $$ n^{\text{th}} $$ term of the series.

The given series is $$ 8x^2 + 4x^3 + 2x^4 + \ldots $$.

We can observe that this series is a geometric series where:
- The first term $$ a = 8x^2 $$
- The common ratio $$ r $$ can be determined by dividing the second term by the first term:

$$
r = \frac{4x^3}{8x^2} = \frac{1}{2}x
$$

The $$ n^{\text{th}} $$ term of a geometric series can be expressed as:

$$
T_n = a \cdot r^{n-1}
$$

Substituting the values of $$ a $$ and $$ r $$:

$$
T_n = 8x^2 \left(\frac{1}{2}x\right)^{n-1}
$$

Thus, the $$ n^{\text{th}} $$ term is:

$$
T_n = 8x^2 \cdot \left(\frac{1}{2}\right)^{n-1} \cdot x^{n-1} = 8x^{n+1} \cdot \left(\frac{1}{2}\right)^{n-1}
$$

### 3.2 For what value(s) of $$ x $$ will the series converge?

A geometric series converges if the absolute value of the common ratio $$ r $$ is less than 1:

$$
\left| r \right| < 1
$$

Substituting the common ratio we found:

$$
\left| \frac{1}{2}x \right| < 1
$$

This simplifies to:

$$
|x| < 2
$$

Thus, the series converges for:

$$
-2 < x < 2
$$

### 3.3 Calculate the sum of the series to infinity if $$ x = \frac{3}{2} $$.

The sum $$ S $$ of an infinite geometric series can be calculated using the formula:

$$
S = \frac{a}{1 - r}
$$

Where:
- $$ a = 8x^2 $$
- $$ r = \frac{1}{2}x $$

Substituting $$ x = \frac{3}{2} $$:

1. Calculate $$ a $$:

$$
a = 8\left(\frac{3}{2}\right)^2 = 8 \cdot \frac{9}{4} = 18
$$

2. Calculate $$ r $$:

$$
r = \frac{1}{2}\left(\frac{3}{2}\right) = \frac{3}{4}
$$

Now, substituting $$ a $$ and $$ r $$ into the sum formula:

$$
S = \frac{18}{1 - \frac{3}{4}} = \frac{18}{\frac{1}{4}} = 18 \cdot 4 = 72
$$

### Final Answers:
- **3.1**: The $$ n^{\text{th}} $$ term is $$ T_n = 8x^{n+1} \cdot \left(\frac{1}{2}\right)^{n-1} $$.
- **3.2**: The series converges for $$ -2 < x < 2 $$.
- **3.3**: The sum of the series to infinity when $$ x = \frac{3}{2} $$ is $$ 72 $$.
- **3.1**: The $$ n^{\text{th}} $$ term is $$ T_n = 8x^{n+1} \cdot \left(\frac{1}{2}\right)^{n-1} $$. - **3.2**: The series converges for $$ -2 < x < 2 $$. - **3.3**: When $$ x = \frac{3}{2} $$, the sum of the series to infinity is $$ 72 $$.

Given the geometric series:
3.1 Determine the term of the series.
3.2 For what value(s) of will the series converge?
3.3 Calculate the sum of the series to infinity if .

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Given the geometric series: 3.1 Determine the term of the series. 3.2 For what value(s) of will the series converge? 3.3 Calculate the sum of the series to infinity if .