A-Level Sequences and series: worked solution

Question

Continuing part (a) with r = 0.6639, build a fresh series using every even-indexed term of that geometric progression (first term b, ratio r). Find the least n for which S differs from the new series' sum to infinity by at least 1.78b.

Step 1: identify the new series. The original geometric progression is b, br, br², br³, br⁴, ... Its even-indexed terms (2nd, 4th, 6th, ...) are br, br³, br⁵, ... Each of these is r² times the one before, so the new series is itself geometric with first term = br and common ratio = r². Step 2: sum the new series to infinity. With r = 0.6639, r² = 0.4408, so |r²| < 1 and the sum to infinity exists. Using first term/(1 - ratio): sum to infinity = br/(1 - r²) = 0.6639b/(1 - 0.6639²) = 0.6639b/(1 - 0.44076) = 0.6639b/0.55924 = 1.18715b Step 3: write S, the sum of n terms of the original progression, and form the difference. S = b(1 - rⁿ)/(1 - r) = b(1 - rⁿ)/(1 - 0.6639) = b(1 - rⁿ)/0.3361 = 2.97531b(1 - rⁿ) S - br/(1 - r²) = 2.97531b - 2.97531b rⁿ - 1.18715b = b(1.78815 - 2.97531rⁿ) Since rⁿ > 0 and shrinks as n grows, S is always the larger of the two and this difference is positive and increasing, approaching 1.78815b. Step 4: impose the requirement that the difference is at least 1.78b, and solve for n. 1.78815 - 2.97531rⁿ ≥ 1.78 -2.97531rⁿ ≥ -0.00815 2.97531rⁿ ≤ 0.00815 rⁿ ≤ 0.002740 Take natural logarithms. Since ln r = ln 0.6639 = -0.40962 is negative, dividing by it reverses the inequality: n ln r ≤ ln(0.002740) n ≥ ln(0.002740)/ln(0.6639) = (-5.8999)/(-0.40962) = 14.40 Step 5: n must be a whole number, so test the integers either side of 14.40. n = 14: difference = b(1.78815 - 2.97531 × 0.6639¹⁴) = 1.7785b, and 1.7785 < 1.78, so n = 14 fails. n = 15: difference = b(1.78815 - 2.97531 × 0.6639¹⁵) = 1.7818b, and 1.7818 ≥ 1.78, so n = 15 works. The least such n is 15.