3.1 Consider \( \sum_{k=0}^{p}\left(3^{5-k}\right)=\frac{1093}{3} \) 3.1.1 Write down the first three terms of the series. 3.1.2 Does the series converge? Motivate your answer.

To solve the problem, let's break it down into the two parts as specified. ### 3.1.1 Write down the first three terms of the series. The series given is: \[ \sum_{k=0}^{p}\left(3^{5-k}\right) \] To find the first three terms, we will substitute \( k = 0, 1, 2 \) into the expression \( 3^{5-k} \). 1. For \( k = 0 \): \[ 3^{5-0} = 3^5 = 243 \] 2. For \( k = 1 \): \[ 3^{5-1} = 3^4 = 81 \] 3. For \( k = 2 \): \[ 3^{5-2} = 3^3 = 27 \] Thus, the first three terms of the series are: \[ 243, 81, 27 \] ### 3.1.2 Does the series converge? Motivate your answer. To determine if the series converges, we need to analyze the nature of the series. The series is a finite sum since it goes from \( k = 0 \) to \( k = p \). 1. **Finite Series**: A finite series always converges because it has a limited number of terms. In this case, the series is defined for \( k \) from \( 0 \) to \( p \), which means it will have \( p + 1 \) terms. 2. **Behavior of Terms**: Each term in the series is of the form \( 3^{5-k} \). As \( k \) increases, the terms decrease exponentially. However, since we are summing a finite number of terms, the series will have a finite sum. Therefore, the series converges because it is a finite sum of terms. ### Conclusion - The first three terms of the series are \( 243, 81, 27 \). - The series converges because it is a finite series.