4. Consider the geometric sequence: \( 8 x^{2} ; 4 x^{3} ; 2 x^{4} ; \ldots \) (a) Show that the general term is given by \( \mathrm{T}_{n}=2^{4-n} . x^{1+n} \) (b) Calculate the sum of the series to infinity if \( x=\frac{3}{2} \).

To tackle the problem, let's start by understanding the geometric sequence given: \(8x^2, 4x^3, 2x^4, \ldots\). **(a)** We can observe the pattern in the sequence: - The first term (\(T_1\)) is \(8x^2 = 2^3 \cdot x^2\). - The second term (\(T_2\)) is \(4x^3 = 2^2 \cdot x^3\). - The third term (\(T_3\)) is \(2x^4 = 2^1 \cdot x^4\). From this, we notice that each term can be expressed as \(T_n = 2^{4-n} \cdot x^{n+1}\). This matches the form we are required to show: \[ T_n = 2^{4-n} \cdot x^{n+1} \] This can be seen by substituting \(n = 1, 2, 3\) and verifying that the initial terms align. **(b)** The series converges to infinity when the ratio \(r\) of consecutive terms satisfies \(|r| < 1\). To find \(r\): \[ \text{The first term} = T_1 = 8x^2 = 8 \cdot \left(\frac{3}{2}\right)^2 = 8 \cdot \frac{9}{4} = 18. \] The second term: \[ T_2 = 4x^3 = 4 \cdot \left(\frac{3}{2}\right)^3 = 4 \cdot \frac{27}{8} = \frac{27}{2}. \] To find \(r\): \[ r = \frac{T_2}{T_1} = \frac{\frac{27}{2}}{18} = \frac{3}{4}. \] Now, the sum \(S\) of an infinite geometric series can be calculated using the formula: \[ S = \frac{T_1}{1 - r}, \] where \(T_1 = 18\) and \(r = \frac{3}{4}\): \[ S = \frac{18}{1 - \frac{3}{4}} = \frac{18}{\frac{1}{4}} = 18 \times 4 = 72. \] Thus, the sum of the series to infinity when \(x = \frac{3}{2}\) is \(72\).