Mathematics/ Investigation NSC - Grade 12 NW/ Feb 2025 3 PART A INVESTIGATING INFINITE GEOMETRIC SERIES RESOURCES: CALCULATOR, A4 PAPER. WHAT IS THE SUM OF AN INFINITE GEOMETRIC SERIES? QUESTION 1 1.1 Given: \( \mathrm{S}_{n}=\frac{a\left(r^{n}-1\right)}{r-1} \) 1.1.1 Write \( S_{n} \) in expanded form (1) 1.1.2 Which term in 1.1 .1 is affected by the number of terms in the series? (1) Consider \( r^{n} \) (write answer in scientific notation form where necessary) 1.2 CASE 1: \( r>1 \) 1.2.1 If \( r=1,001 \) determine the values of (a) \( r^{200} \) and \( S_{200} \) (b) \( r^{20000} \) and \( \mathrm{S}_{20000} \) (2) 1.2.2 Determine \( r^{200} \) and \( \mathrm{S}_{200} \) if: (a) \( r=\frac{5}{2} \) (b) \( r=3 \) (2) 1.2.3 Make a conjecture with regard to \( r^{n} \) and \( S_{n} \) as \( n \rightarrow \infty \) (2) 1.3 CASE 2: \( r<-1 \) 1.3.1 If \( r=-1,001 \), determine the value of \( r^{n} \) if (a) \( n=9999 \) (b) \( n=10000 \)

Below is one acceptable worked‐out solution. (Some numerical answers have been rounded to an appropriate number of significant figures.) It is assumed that the first term is a (and in parts where a number is not specified we assume a remains as a constant, so in many cases you may take a = 1 for numerical evaluation). ────────────────────────────── 1.1 Expanded Form The formula for the sum of a geometric series is   Sₙ = a (rⁿ – 1)/(r – 1). 1.1.1 To expand Sₙ we note that   a (rⁿ – 1) = a (r – 1)(1 + r + r² + ⋯ + rⁿ⁻¹). Thus, equivalently,   Sₙ = a + a·r + a·r² + … + a·rⁿ⁻¹. 1.1.2 In the expanded form, the term that “knows” how many terms are there is the last term, namely a·rⁿ⁻¹; equivalently, in the compact formula the factor rⁿ (in the numerator) is affected by the number of terms n. ────────────────────────────── 1.2 CASE 1: r > 1 For the following, when a numerical value is required, we often assume a = 1. 1.2.1 Given r = 1.001  (a) For n = 200:   • Compute r²⁰⁰ = (1.001)²⁰⁰.    A way to approximate is to use logarithms:     ln[(1.001)²⁰⁰] = 200·ln(1.001) ≈ 200·0.001 = 0.2.    So, r²⁰⁰ ≈ e^(0.2) ≈ 1.2214  (≈ 1.22×10⁰ in sci. notation).   • Then     S₂₀₀ = (r²⁰⁰ – 1)/(1.001 – 1) ≈ (1.2214 – 1)/0.001 = 0.2214/0.001 = 221.4.  (b) For n = 20,000:   • First, r^(20000) = (1.001)^(20000).    Again, ln[(1.001)^(20000)] = 20000·ln(1.001) ≈ 20000·0.001 = 20.    Thus, r^(20000) ≈ e^(20) ≈ 4.85×10⁸.   • Then,    S₂₀₀₀₀ = (r^(20000) – 1)/(1.001 – 1) ≈ (4.85×10⁸ – 1)/0.001.    Since 1 is negligible compared to 4.85×10⁸,    S₂₀₀₀₀ ≈ 4.85×10⁸/0.001 = 4.85×10¹¹. ────────────────────────────── 1.2.2 Determine r²⁰⁰ and S₂₀₀ if  (a) r = 5/2 = 2.5   • To compute r²⁰⁰ = (2.5)²⁰⁰, use logarithms:    log₁₀(2.5) ≈ 0.39794 so that    log₁₀((2.5)²⁰⁰) ≈ 200 × 0.39794 = 79.588.    Thus, (2.5)²⁰⁰ ≈ 10^(79.588) which can be written approximately as ≈ 3.88×10⁷⁹.   • Then     S₂₀₀ = ((2.5)²⁰⁰ – 1)/(2.5 – 1) ≈ (3.88×10⁷⁹)/(1.5) ≈ 2.59×10⁷⁹.    (The “–1” is negligible in comparison with 3.88×10⁷⁹.)  (b) r = 3   • Similarly, r²⁰⁰ = 3²⁰⁰.    Since log₁₀(3) ≈ 0.4771, we have    log₁₀(3²⁰⁰) ≈ 200 × 0.4771 = 95.42.    Thus, 3²⁰⁰ ≈ 10^(95.42) ≈ 2.63×10⁹⁵.   • Then,    S₂₀₀ = (3²⁰⁰ – 1)/(3 – 1) ≈ (2.63×10⁹⁵)/(2) ≈ 1.315×10⁹⁵. ────────────────────────────── 1.2.3 Conjecture as n → ∞ for r > 1 Notice that as n increases without bound:   • rⁿ grows very rapidly (in fact, rⁿ → ∞ when r > 1).   • Consequently, Sₙ = a (rⁿ – 1)/(r – 1) also increases without bound (approaching infinity). Conjecture: For any r > 1, as n → ∞, both rⁿ and Sₙ tend to infinity. ────────────────────────────── 1.3 CASE 2: r < –1 1.3.1 Given r = –1.001, determine rⁿ for:  (a) n = 9999   Since n is odd, the sign will be negative.   Compute |r|ⁿ = (1.001)^(9999).   Using ln approximation: ln(1.001) ≈ 0.001 so that    ln[(1.001)^(9999)] ≈ 9999 × 0.001 = 9.999.   Thus, (1.001)^(9999) ≈ e^(9.999) ≈ 2.20×10⁴.   Therefore, r^(9999) ≈ –2.20×10⁴.  (b) n = 10000   Now n is even so the negative sign is lost.   |r|^(10000) = (1.001)^(10000).   Approximate: ln[(1.001)^(10000)] ≈ 10000 × 0.001 = 10, so that   (1.001)^(10000) ≈ e^(10) ≈ 2.20×10⁴.   Thus, r^(10000) ≈ +2.20×10⁴. ────────────────────────────── Note: • When writing answers in scientific notation, one might write, for example, 1.2214 as 1.22×10⁰ and 4.85×10⁸ as 4.85×10⁸. • Where rounding is involved, the level of accuracy depends on your instructions; the above provides one acceptable level. This completes one acceptable investigation for this part of the investigation into infinite geometric series.