3.1 Consider \( \sum_{k=0}^{n}\left(3^{s-k}\right)=\frac{1093}{3} \) 3.1.1 \( \quad \) Write down the first three terms of the series. 3.1.2 \( \begin{array}{ll}\text { 3.1.3 } & \text { Does the series converge? Motivate your answer. }\end{array} \).

To solve the problem, let's break it down step by step. ### 3.1.1: Write down the first three terms of the series. The series given is: \[ \sum_{k=0}^{n}\left(3^{s-k}\right) \] The first three terms of this series can be found by substituting \( k = 0, 1, 2 \): 1. For \( k = 0 \): \[ 3^{s-0} = 3^s \] 2. For \( k = 1 \): \[ 3^{s-1} = \frac{3^s}{3} = 3^{s-1} \] 3. For \( k = 2 \): \[ 3^{s-2} = \frac{3^{s-1}}{3} = \frac{3^s}{3^2} = 3^{s-2} \] Thus, the first three terms of the series are: \[ 3^s, \quad 3^{s-1}, \quad 3^{s-2} \] ### 3.1.2: Does the series converge? Motivate your answer. To determine whether the series converges, we need to analyze the behavior of the terms as \( n \) approaches infinity. The series can be rewritten as: \[ \sum_{k=0}^{n} 3^{s-k} = 3^s + 3^{s-1} + 3^{s-2} + \ldots + 3^{s-n} \] This is a geometric series with the first term \( a = 3^s \) and the common ratio \( r = \frac{1}{3} \). A geometric series converges if the absolute value of the common ratio \( |r| < 1 \). In this case, since \( r = \frac{1}{3} \), we have: \[ |r| = \frac{1}{3} < 1 \] Thus, the series converges. ### Conclusion 1. The first three terms of the series are \( 3^s, 3^{s-1}, 3^{s-2} \). 2. The series converges because it is a geometric series with a common ratio \( |r| < 1 \).