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? 2.1.2 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 \( \frac{49}{2} \).

To determine the convergence of the series \( \sum_{k=0}^{\infty} x^{2}(x-3)^{k} \), we recognize it as a geometric series where the common ratio is \( (x-3) \). For a geometric series to converge, the absolute value of the common ratio must be less than 1, leading to the condition \( |x-3| < 1 \). Therefore, the series converges for \( 2 < x < 4 \). Now, substituting \( x = \frac{5}{2} \) into the series, we have \( \sum_{k=0}^{\infty} \left(\frac{5}{2}\right)^{2} \left(\frac{5}{2} - 3\right)^{k} = \sum_{k=0}^{\infty} \frac{25}{4} \left(-\frac{1}{2}\right)^{k} \). The first term is \( \frac{25}{4} \) and the common ratio \( r = -\frac{1}{2} \), and since \( |r| < 1 \), we can sum the series using the formula \( S = \frac{a}{1 - r} \), leading to \( S = \frac{\frac{25}{4}}{1 - (-\frac{1}{2})} = \frac{\frac{25}{4}}{\frac{3}{2}} = \frac{25}{6} \). To find \( x \) such that the sum to infinity is \( \frac{49}{2} \), we set up the equation \( \frac{x^{2}}{1 - (x-3)} = \frac{49}{2} \). This simplifies to \( \frac{x^{2}}{4 - x} = \frac{49}{2} \). Cross-multiplying gives \( 2x^{2} = 49(4 - x) \), leading to \( 2x^{2} + 49x - 196 = 0 \). Solving this quadratic equation via the quadratic formula will yield the values of \( x \) for which the sum is \( \frac{49}{2} \).