2.1. Consider the series: \( 125(x-3)+25(x-3)^{2}+5(x-3)^{3}+\cdots \) 2.1.1 For which value(s) of \( x \) will the series be conivergent? 2.1.2 Determine the sum to infinity of the series if \( x=7 \) 2.2. The sum of the first \( n \) terms of a series is given by \( S_{n}=2\left(3^{n}-1\right) \) 2.2.1. Determine the sum of the first 12 terms. 2.2.2. Determine the \( 12^{\text {th }} \) terme \( ~ \)

2.1. We are given the series   125(x – 3) + 25(x – 3)² + 5(x – 3)³ + ⋯ Notice that the ratio between consecutive terms is   [25(x – 3)²]⁄[125(x – 3)] = (1/5)(x – 3) and similarly for the other terms. Hence, the series is geometric with   First term, a = 125(x – 3)   Common ratio, r = (x – 3)/5 2.1.1. A geometric series converges when |r| < 1. Therefore, we require   |(x – 3)/5| < 1  ⟹  |x – 3| < 5 This inequality is satisfied when   –5 < x – 3 < 5  ⟹  –2 < x < 8 Thus, the series converges for x between –2 and 8. 2.1.2. To find the sum to infinity when x = 7, first substitute x = 7 into the formulas for a and r:   a = 125(7 – 3) = 125·4 = 500   r = (7 – 3)/5 = 4/5 Since |4/5| < 1, the sum to infinity is given by   S∞ = a/(1 – r) = 500/(1 – 4/5) = 500/(1/5) = 500·5 = 2500. 2.2. We are given that the sum of the first n terms of a series is   Sₙ = 2(3ⁿ – 1). 2.2.1. To find the sum of the first 12 terms, substitute n = 12:   S₁₂ = 2(3¹² – 1). We compute 3¹². Since 3¹² = 531441, it follows that   S₁₂ = 2(531441 – 1) = 2(531440) = 1,062,880. 2.2.2. The 12th term, a₁₂, is the difference between S₁₂ and S₁₁ (since Sₙ gives the sum of the first n terms):   a₁₂ = S₁₂ – S₁₁. Now, compute S₁₁ by substituting n = 11:   S₁₁ = 2(3¹¹ – 1). Since 3¹¹ = 177147, we have   S₁₁ = 2(177147 – 1) = 2(177146) = 354292. Thus,   a₁₂ = 1,062,880 – 354292 = 708588. Alternatively, we can derive a formula for the nth term. Since   aₙ = Sₙ – Sₙ₋₁ = 2(3ⁿ – 1) – 2(3ⁿ⁻¹ – 1) = 2(3ⁿ – 3ⁿ⁻¹) = 2·3ⁿ⁻¹(3 – 1),   aₙ = 4·3ⁿ⁻¹. Then the 12th term is   a₁₂ = 4·3¹¹ = 4·177147 = 708588. Summary of Answers:  • 2.1.1. The series converges for –2 < x < 8.  • 2.1.2. When x = 7, the sum to infinity is 2500.  • 2.2.1. The sum of the first 12 terms is 1,062,880.  • 2.2.2. The 12th term is 708588.