Dotermine the valuen of \( x \) for which cach of the following geometric series converges \( \begin{array}{ll}\text { (1) } x+2 x^{2}+4 x^{3}+\ldots & \text { (2) } 1-x+x^{2}-\ldots \\ \text { (3) }(x+1)+(x+1)^{2}+(x+1)^{3}+\ldots & \text { (4) } 2(1-x)+2(1-x)^{2}+2(1-x)^{3}+\ldots \\ \text { (5) }(2 x-5)+(2 x-5)^{2}+(2 x-5)^{3}+\ldots & \text { (6) } x(7-2 x)^{4}-x(7-2 x)^{5}+x(7-2 \\ \text { (7) }(3 x-2)^{3}-(3 x-2)^{4}+(3 x-2)^{5}-\ldots & \text { (8) }(2-4 x)+2(2 x-1)^{2}-2(2 x-1)^{3} \\ \text { (he first two torms of a geometric serics are } x+4 \text { and } x^{2}-16,\end{array} \)

To determine when these geometric series converge, we utilize the formula for convergence of a geometric series, which states that a series converges if the absolute value of the common ratio \( r \) is less than 1, written as \( |r| < 1 \). 1. For the series \( x + 2x^2 + 4x^3 + \ldots \), the first term \( a = x \) and the common ratio \( r = \frac{2x^2}{x} = 2x \). Therefore, the series converges if \( |2x| < 1 \) leading to \( -\frac{1}{2} < x < \frac{1}{2} \). 2. The series \( 1 - x + x^2 - \ldots \) has \( a = 1 \) and \( r = -x \). This series is convergent if \( |-x| < 1 \) which simplifies to \( -1 < x < 1 \). 3. For the series \( (x+1) + (x+1)^2 + (x+1)^3 + \ldots \), \( a = (x+1) \) and \( r = (x+1) \). Thus, it converges when \( |x+1| < 1 \), which means \( -2 < x < 0 \). 4. In the series \( 2(1-x) + 2(1-x)^2 + 2(1-x)^3 + \ldots \), we have \( a = 2(1-x) \) and \( r = (1-x) \). The series converges if \( |1-x| < 1 \), leading to \( 0 < x < 2 \). 5. For the series \( (2x-5) + (2x-5)^2 + (2x-5)^3 + \ldots \), \( a = (2x-5) \) and \( r = (2x-5) \). The series converges if \( |2x-5| < 1 \). This gives the inequalities \( 4 < 2x < 6 \), leading to \( 2 < x < 3 \). 6. In the expression \( x(7-2x)^{4} - x(7-2x)^{5} + x(7-2x)^{6} - \ldots \), the first term is \( a = x(7-2x)^{4} \) and \( r = -(7-2x) \). The condition for convergence is \( |7-2x| < 1 \), giving \( 3 < x < 4 \). 7. The series \( (3x - 2)^{3} - (3x - 2)^{4} + (3x - 2)^{5} - \ldots \) has \( a = (3x - 2)^{3} \) and \( r = -(3x - 2) \). Thus, it converges if \( |3x - 2| < 1 \), leading to \( \frac{1}{3} < x < 1 \). 8. Lastly, in the series \( (2-4x) + 2(2x-1)^{2} - 2(2x-1)^{3} + \ldots \), we find \( a = 2-4x \) and the common ratio based on the subsequent terms. It converges if \( |2x-1| < 1 \), resulting in \( 0 < x < 1 \). By applying these conditions across each series, we can clearly see the ranges of \( x \) for which they converge.