iven: \( 25\left(\frac{1-1}{3}\right)+5\left(\frac{1-t}{3}\right)^{2}+\left(\frac{1-1}{3}\right)^{3}+\ldots . . . . . \). 4.1 For which value(s) of \( t \) will the series converga

To determine for which values of \( t \) the series converges, observe that the series can be simplified. The first term \( 25\left(\frac{1-1}{3}\right) = 25(0) = 0 \), and the last term is also zero, rendering the terms involving \( \frac{1-1}{3} \) irrelevant. The second term \( 5 \left(\frac{1-t}{3}\right)^{2} \) indicates that the series is primarily focused on the second factor. For the series to converge, the general term \( \left(\frac{1-t}{3}\right)^{n} \) must go to zero as \( n \) approaches infinity, which occurs when the absolute value is less than 1. Thus, we require: \[ \left| \frac{1-t}{3} \right| < 1 \] This means: \[ -1 < \frac{1-t}{3} < 1 \] Multiplying through by 3 gives: \[ -3 < 1 - t < 3 \] From this, we can isolate \( t \): 1. For \( 1 - t > -3 \), we get \( t < 4 \). 2. For \( 1 - t < 3 \), we get \( t > -2 \). Putting these inequalities together, the series converges for: \[ -2 < t < 4 \]