$$ 3.2 \begin{array}{l} ext { The first } 3 ext { terms of an infinite geometric series are given: } \ (x+1)+2(x+1)^{2}+4(x+1)^{3}+\ldots .\end{array} $$ $$ \begin{array}{ll}3.2 .1 & ext { For which values of } x ext { will the series converge? } \ 3.2 .2 & ext { If } x=-\frac{3}{4} ext {, determine the numerical value of the first term. } \ \begin{array}{ll}3.2 .3 & ext { Write the series in sigma notation. } \ 3.2 .4 & ext { Calculate the sum to infinity of the series. }\end{array}\end{array} $$.

Question

$$ 3.2 \begin{array}{l}	ext { The first } 3 	ext { terms of an infinite geometric series are given: } \ (x+1)+2(x+1)^{2}+4(x+1)^{3}+\ldots .\end{array} $$ $$ \begin{array}{ll}3.2 .1 & 	ext { For which values of } x 	ext { will the series converge? } \ 3.2 .2 & 	ext { If } x=-\frac{3}{4} 	ext {, determine the numerical value of the first term. } \ \begin{array}{ll}3.2 .3 & 	ext { Write the series in sigma notation. } \ 3.2 .4 & 	ext { Calculate the sum to infinity of the series. }\end{array}\end{array} $$.

UpStudy AI · Accepted Answer

Let's analyze the given infinite geometric series and solve each part step by step.

**Given Series:**
$$
(x + 1) + 2(x + 1)^2 + 4(x + 1)^3 + \ldots
$$

This series is geometric, where each term after the first is obtained by multiplying the previous term by a common ratio, $$ r $$.

---

### **3.2.1 For which values of $$ x $$ will the series converge?**

**Solution:**

An infinite geometric series converges if and only if the absolute value of the common ratio is less than 1, i.e., $$ |r| < 1 $$.

**Finding the Common Ratio ($$ r $$):**
- First term ($$ a $$) = $$ (x + 1) $$
- Second term = $$ 2(x + 1)^2 $$

$$
r = \frac{\text{Second term}}{\text{First term}} = \frac{2(x + 1)^2}{x + 1} = 2(x + 1)
$$

**Setting the condition for convergence:**

$$
|2(x + 1)| < 1
$$

$$
|x + 1| < \frac{1}{2}
$$

**Solving the inequality:**

$$
-\frac{1}{2} < x + 1 < \frac{1}{2}
$$

$$
-1.5 < x < -0.5
$$

**Answer:**
The series converges for all real numbers $$ x $$ such that:
$$
- \frac{3}{2} < x < -\frac{1}{2}
$$

---

### **3.2.2 If $$ x = -\frac{3}{4} $$, determine the numerical value of the first term.**

**Solution:**

First term ($$ a $$) = $$ (x + 1) $$

Substitute $$ x = -\frac{3}{4} $$:

$$
a = -\frac{3}{4} + 1 = \frac{1}{4}
$$

**Answer:**
The first term is $$ \frac{1}{4} $$.

---

### **3.2.3 Write the series in sigma notation.**

**Solution:**

The given series:
$$
(x + 1) + 2(x + 1)^2 + 4(x + 1)^3 + \ldots
$$

Observe that the coefficients are powers of 2:
$$
1, 2, 4, 8, \ldots = 2^{0}, 2^{1}, 2^{2}, 2^{3}, \ldots
$$

Each term can be expressed as:
$$
2^{n-1}(x + 1)^n \quad \text{for} \quad n = 1, 2, 3, \ldots
$$

**Sigma Notation:**
$$
\sum_{n=1}^{\infty} 2^{n-1} (x + 1)^{n}
$$

**Answer:**
The series in sigma notation is
$$
\sum_{n=1}^{\infty} 2^{\,n-1}\, (x + 1)^{\,n}
$$

---

### **3.2.4 Calculate the sum to infinity of the series.**

**Solution:**

For a convergent geometric series, the sum to infinity ($$ S $$) is given by:
$$
S = \frac{a}{1 - r}
$$
where $$ a $$ is the first term and $$ r $$ is the common ratio.

**Given:**
- $$ a = (x + 1) $$
- $$ r = 2(x + 1) $$

**Substitute into the formula:**
$$
S = \frac{x + 1}{1 - 2(x + 1)} = \frac{x + 1}{1 - 2x - 2} = \frac{x + 1}{-2x - 1}
$$

For simplicity, this can also be written as:
$$
S = -\frac{x + 1}{2x + 1}
$$

**Note:** This formula is valid only for values of $$ x $$ that ensure $$ |r| < 1 $$, i.e., $$ -\frac{3}{2} < x < -\frac{1}{2} $$ as determined in part 3.2.1.

**Answer:**
The sum to infinity is
$$
S = \frac{x + 1}{\,1 - 2(x + 1)\,} = \frac{x + 1}{-\,2x\;-\;1\,}
$$

---

**Summary of Answers:**

1. **Convergence:** $$ -\frac{3}{2} < x < -\frac{1}{2} $$
2. **First Term When $$ x = -\frac{3}{4} $$:** $$ \frac{1}{4} $$
3. **Sigma Notation:** $$ \displaystyle \sum_{n=1}^{\infty} 2^{\,n-1}\, (x + 1)^{\,n} $$
4. **Sum to Infinity:** $$ S = \frac{x + 1}{-2x - 1} $$
**Summary of Answers:** 1. **Convergence:** $$ -\frac{3}{2} < x < -\frac{1}{2} $$ 2. **First Term When $$ x = -\frac{3}{4} $$:** $$ \frac{1}{4} $$ 3. **Sigma Notation:** $$ \sum_{n=1}^{\infty} 2^{n-1} (x + 1)^n $$ 4. **Sum to Infinity:** $$ S = \frac{x + 1}{-2x - 1} $$

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