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 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)]. Step 1. Write out the general term:   aₖ = 4 · 3^(2–k). For k = 1, 2, 3,… the terms are:   a₁ = 4 · 3^(2–1) = 4 · 3 = 12,   a₂ = 4 · 3^(2–2) = 4 · 1 = 4,   a₃ = 4 · 3^(2–3) = 4 · 3^(–1) = 4/3,   … and so on. 2.1 Convergence To decide convergence for a geometric series, we look at the common ratio. Notice that the term 3^(2–k) can be rewritten as 3² · 3^(–k) = 9 · 3^(–k), so the series becomes   4 · 9 · ∑ₖ₌₁∞ (1/3)ᵏ = 36 ∑ₖ₌₁∞ (1/3)ᵏ. This is a geometric series with common ratio r = 1/3. Since |r| = 1/3 < 1, the series converges. Thus, the series is convergent. 2.2 Finding p such that   ∑ₖ₌ₚ∞ [4 · 3^(2–k)] = 2/9. For a geometric series starting at k = p, the sum is given by   S = aₚ/(1 – r) where   aₚ = 4 · 3^(2–p) and r = 1/3. Substitute these into the sum formula:   S = [4 · 3^(2–p)] / (1 – 1/3) = [4 · 3^(2–p)] / (2/3) = 4 · 3^(2–p) · (3/2) = 6 · 3^(2–p). We are given that S = 2/9, so   6 · 3^(2–p) = 2/9. Solve for 3^(2–p) by dividing both sides by 6:   3^(2–p) = (2/9) / 6 = 2/(9·6) = 2/54 = 1/27. Notice that 1/27 can be written as 3^(–3), because 27 = 3³. Thus, we equate the exponents:   3^(2–p) = 3^(–3)  ⟹  2 – p = –3. Solve for p:   2 – p = –3  ⟹  –p = –5  ⟹  p = 5. Final Answers:  2.1 The series is convergent because the common ratio is 1/3 (with |1/3| < 1).  2.2 The value of p such that ∑ₖ₌ₚ∞ [4 · 3^(2–k)] = 2/9 is p = 5.