QUESTION 2 Consider the series $$ \sum_{t=0}^{\infty} x^{2}(x-3)^{\prime} $$ 2.1.1 Determine the first three terms of the series. 2.1.2 For which values of $$ x $$ will this series converge? 2.1.3 If $$ x=\frac{5}{2} $$, calculate the sum to infinity of the series.

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**Step 1. Write the series in expanded form.**

The series is given by
$$
\sum_{t=0}^{\infty} x^{2}(x-3)^t.
$$
This means the general term (the $$t$$th term) is
$$
a_t = x^2 (x-3)^t.
$$

---

**Step 2. Determine the first three terms.**

For $$t = 0$$:
$$
a_0 = x^2 (x-3)^0 = x^2.
$$

For $$t = 1$$:
$$
a_1 = x^2 (x-3)^1 = x^2 (x-3).
$$

For $$t = 2$$:
$$
a_2 = x^2 (x-3)^2.
$$

Thus, the first three terms of the series are
$$
x^2,\quad x^2 (x-3),\quad x^2 (x-3)^2.
$$

---

**Step 3. Determine the values of $$x$$ for which the series converges.**

The series is a geometric series with the first term $$a = x^2$$ and common ratio $$r = x-3$$.

A geometric series converges if and only if 
$$
|r| < 1.
$$
Here, this condition becomes
$$
|x-3| < 1.
$$

To solve the inequality:
$$
-1 < x-3 < 1,
$$
add 3 to each part:
$$
2 < x < 4.
$$

Thus, the series converges for 
$$
x \in (2, 4).
$$

---

**Step 4. Calculate the sum to infinity when $$x = \frac{5}{2}$$.**

When $$x = \frac{5}{2}$$:
- First, calculate the common ratio:
  $$
  r = x-3 = \frac{5}{2} - 3 = \frac{5-6}{2} = -\frac{1}{2}.
  $$

- The first term is:
  $$
  a = x^2 = \left(\frac{5}{2}\right)^2 = \frac{25}{4}.
  $$

Since $$|r| = \frac{1}{2} < 1$$, the sum to infinity of a geometric series is given by
$$
S_\infty = \frac{a}{1-r}.
$$

Substitute the values:
$$
S_\infty = \frac{\frac{25}{4}}{1 - \left(-\frac{1}{2}\right)} = \frac{\frac{25}{4}}{1 + \frac{1}{2}}.
$$

Simplify the denominator:
$$
1 + \frac{1}{2} = \frac{3}{2}.
$$

Thus,
$$
S_\infty = \frac{\frac{25}{4}}{\frac{3}{2}} = \frac{25}{4} \times \frac{2}{3} = \frac{25}{6}.
$$

---

**Final Answers:**

- The first three terms of the series are:
  $$
  x^2,\quad x^2 (x-3),\quad x^2 (x-3)^2.
  $$

- The series converges for:
  $$
  x \in (2, 4).
  $$

- For $$ x=\frac{5}{2} $$, the sum to infinity is:
  $$
  \frac{25}{6}.
  $$
**First Three Terms:** $$ x^2, \ x^2(x - 3), \ x^2(x - 3)^2 $$ **Convergence Range:** $$ x $$ between 2 and 4 **Sum When $$ x = \frac{5}{2} $$:** $$ \frac{25}{6} $$

QUESTION 2 Consider the series \( \sum_{t=0}^{\infty} x^{2}(x-3)^{\prime} \) 2.1.1 Determine the first three terms of the series. 2.1.2 For which values of \( x \) will this series converge? 2.1.3 If \( x=\frac{5}{2} \), calculate the sum to infinity of the series.

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