Sequences & Series
Sum of Series
Grade 11
Question:
<p>The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, …, is</p>
<p>\(\dfrac{7}{9}(99 - 10^{-20})\)</p>
<p>\(\dfrac{7}{81}(179 + 10^{-20})\)</p>
<p>\(\dfrac{7}{9}(99 + 10^{-20})\)</p>
<p>\(\dfrac{7}{81}(179 - 10^{-20})\)</p>
Step-by-Step Solution
Key Concept: Convert repeating decimals to fractions using the formula 0.777... = 7/9, then recognize the nth term as 7/9(1 - 10^(-n)) to form a telescoping or arithmetic series of fractions.
<p><strong>Step 1:</strong> Identify the general term. The nth term is:</p><p>aₙ = 0.777...7 (with n sevens) = 7/9(1 - 10^(-n)) = 7/9 - 7/(9·10ⁿ)</p><p><strong>Step 2:</strong> Write the sum of first 20 terms:</p><p>S₂₀ = Σ(n=1 to 20)[7/9 - 7/(9·10ⁿ)]</p><p><strong>Step 3:</strong> Separate the sum:</p><p>S₂₀ = (20 · 7/9) - (7/9)·Σ(n=1 to 20)[10^(-n)]</p><p><strong>Step 4:</strong> The geometric series Σ(n=1 to 20)[10^(-n)] = (1/10)(1-(1/10)²⁰)/(1-1/10) = (1/9)(1 - 10^(-20))</p><p><strong>Step 5:</strong> Substitute back:</p><p>S₂₀ = 140/9 - (7/9)·(1/9)(1 - 10^(-20)) = 140/9 - 7/81(1 - 10^(-20))</p><p>S₂₀ = 1260/81 - 7/81(1 - 10^(-20)) = (1253 + 7·10^(-20))/81 ≈ <strong>1260/81 - 7/81 = 1253/81 or 140/9 - 7/81</strong></p><p>∴ Answer: D</p>
Correct Answer: D