QUESTION 2 Consider the series: $$ \sum_{k=0}^{\infty} x^{2}(x-3)^{k} $$ 2.1.1. For which values of $$ x $$ will this series converge? 2., If $$ x=\frac{5}{2} $$, calculate the sum to infinity of the series. 2.1.3. Determine the value of $$ x $$ if the sum to infinity is $$ \frac{49}{2} $$.

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Let's analyze the given series step by step:

$$ \sum_{k=0}^{\infty} x^{2}(x-3)^{k} $$

This is a geometric series where each term is of the form $$ x^2 (x-3)^k $$. For a geometric series $$ \sum_{k=0}^{\infty} ar^k $$ to converge, the common ratio $$ r $$ must satisfy $$ |r| < 1 $$.

### 2.1.1. **Values of $$ x $$ for Convergence**

- **Common Ratio Identification**:
  - Here, $$ r = x - 3 $$.

- **Convergence Condition**:
  $$
  |x - 3| < 1
  $$

- **Solution**:
  $$
  -1 < x - 3 < 1 \
  2 < x < 4
  $$

**Therefore, the series converges for all real numbers $$ x $$ such that:**
$$
2 < x < 4
$$

---

### 2.1.2. **Sum to Infinity When $$ x = \frac{5}{2} $$**

- **Given**:
  $$
  x = \frac{5}{2} = 2.5
  $$

- **Verification of Convergence**:
  $$
  |2.5 - 3| = 0.5 < 1 \quad \text{(Condition satisfied)}
  $$

- **Sum of the Series**:
  The sum of an infinite geometric series is given by:
  $$
  S = \frac{a}{1 - r}
  $$
  where $$ a = x^2 $$ and $$ r = x - 3 $$.

- **Substituting Values**:
  $$
  S = \frac{(2.5)^2}{1 - (2.5 - 3)} = \frac{6.25}{1 - (-0.5)} = \frac{6.25}{1.5} = \frac{25}{6}
  $$

**Therefore, the sum to infinity when $$ x = \frac{5}{2} $$ is:**
$$
\frac{25}{6}
$$

---

### 2.1.3. **Value of $$ x $$ When the Sum to Infinity is $$ \frac{49}{2} $$**

- **Given**:
  $$
  S = \frac{49}{2}
  $$

- **Using the Sum Formula**:
  $$
  \frac{x^2}{1 - (x - 3)} = \frac{49}{2} \
  \frac{x^2}{4 - x} = \frac{49}{2}
  $$

- **Solving for $$ x $$**:
  $$
  2x^2 = 49(4 - x) \
  2x^2 = 196 - 49x \
  2x^2 + 49x - 196 = 0
  $$

- **Applying the Quadratic Formula**:
  $$
  x = \frac{-49 \pm \sqrt{49^2 + 4 \cdot 2 \cdot 196}}{2 \cdot 2} \
  x = \frac{-49 \pm \sqrt{2401 + 1568}}{4} \
  x = \frac{-49 \pm \sqrt{3969}}{4} \
  x = \frac{-49 \pm 63}{4}
  $$

- **Possible Solutions**:
  $$
  x = \frac{14}{4} = 3.5 \quad \text{or} \quad x = \frac{-112}{4} = -28
  $$

- **Verification of Convergence**:
  - For $$ x = 3.5 $$:
    $$
    |3.5 - 3| = 0.5 < 1 \quad \text{(Valid)}
    $$
  - For $$ x = -28 $$:
    $$
    |-28 - 3| = 31 > 1 \quad \text{(Invalid)}
    $$

**Only $$ x = 3.5 $$ satisfies the convergence condition.**

**Therefore, the value of $$ x $$ is:**
$$
x = \frac{7}{2} \quad \text{or} \quad 3.5
$$

---

**Summary of Answers:**

1. **Convergence**: For all real numbers $$ x $$ with $$ 2 < x < 4 $$.

2. **Sum when $$ x = \frac{5}{2} $$**: $$ \frac{25}{6} $$.

3. **Value of $$ x $$ for sum $$ \frac{49}{2} $$**: $$ \frac{7}{2} $$.
The series $$ \sum_{k=0}^{\infty} x^{2}(x-3)^{k} $$ converges for $$ 2 < x < 4 $$. When $$ x = \frac{5}{2} $$, the sum to infinity is $$ \frac{25}{6} $$. If the sum to infinity is $$ \frac{49}{2} $$, then $$ x = \frac{7}{2} $$.

QUESTION 2 Consider the series: \( \sum_{k=0}^{\infty} x^{2}(x-3)^{k} \) 2.1.1. For which values of \( x \) will this series converge? 2., If \( x=\frac{5}{2} \), calculate the sum to infinity of the series. 2.1.3. Determine the value of \( x \) if the sum to infinity is \( \frac{49}{2} \).

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