APPLICATION QUESTION 2 Given that: \( \sum_{k=1}^{\infty} 4.3^{2-k} \) 2.1 Is the above geometric series is convergent or divergent. Justify your answer 2025 2.2 If further given that \( \sum_{k=p}^{\infty} 4.3^{2-k}=\frac{2}{9} \), determine the value of \( p \).

We are given the series   ∑ₖ₌₁^∞ 4·3^(2–k) and we need to address two parts. ────────────────────────────── Part 2.1: Convergence of the Series A geometric series has the form   a + ar + ar² + ar³ + ⋯ and it converges if |r| < 1. Step 1. Identify the first term and common ratio. For k = 1, the first term is:   a = 4·3^(2–1) = 4·3¹ = 12. To find the common ratio r, observe how the term changes when k increases by 1:   Term for k:  4·3^(2–k)   Term for k+1: 4·3^(2–(k+1)) = 4·3^(2–k–1) = 4·3^(2–k)·3^(–1). Thus, the ratio is:   r = [4·3^(2–k)·3^(–1)] ÷ [4·3^(2–k)] = 3^(–1) = 1/3. Step 2. Check the convergence criterion. Since |r| = |1/3| < 1, the series is convergent. ────────────────────────────── Part 2.2: Finding a Specific Starting Index p We are given that   ∑ₖ₌ₚ^∞ 4·3^(2–k) = 2/9. For a geometric series starting at index p, the first term is:   aₚ = 4·3^(2–p), and the sum of the infinite geometric series is:   Sₚ = aₚ / (1 – r), where r = 1/3. Step 1. Write the expression for the sum from k = p to infinity:   Sₚ = [4·3^(2–p)] / (1 – 1/3). Since 1 – 1/3 = 2/3, we have:   Sₚ = [4·3^(2–p)] / (2/3) = 4·3^(2–p) · (3/2) = 6·3^(2–p). Step 2. Set this equal to 2/9 and solve for p:   6·3^(2–p) = 2/9. Divide both sides by 6:   3^(2–p) = (2/9) / 6 = 2/54 = 1/27. Recall that 1/27 can be written as 3^(–3):   3^(2–p) = 3^(–3). By equating the exponents:   2 – p = –3. Solve for p:   p = 2 + 3 = 5. ────────────────────────────── Final Answers: 2.1 The series is convergent because it is geometric with a common ratio |1/3| < 1. 2.2 The value of p such that ∑ₖ₌ₚ^∞ 4·3^(2–k) equals 2/9 is p = 5.