Sequences & Series
Series Summation
Grade 11
Question:
<p>The sum of 20 terms of the series whose \(r\)th term is given by \(T(n) = (-1)^n \dfrac{n^2 + n + 1}{n!}\) is</p>
<p>\(\dfrac{20}{19!} - 2\)</p>
<p>\(\dfrac{21}{20!} - 1\)</p>
<p>\(\dfrac{21}{20!}\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Decompose the numerator n² + n + 1 into terms that telescope or relate to factorial derivatives. Notice that n² + n + 1 = n(n+1) + 1, allowing us to write Tₙ = (-1)ⁿ[n(n+1)/n! + 1/n!] = (-1)ⁿ[1/(n-1)! + 1/n!].
<p><strong>Step 1:</strong> Decompose the general term.</p><p>Given: Tₙ = (-1)ⁿ(n² + n + 1)/n!</p><p>Rewrite numerator: n² + n + 1 = n(n+1) + 1</p><p>So: Tₙ = (-1)ⁿ[n(n+1)/n! + 1/n!] = (-1)ⁿ[1/(n-1)! + 1/n!]</p><p><strong>Step 2:</strong> Separate into two sums.</p><p>S₂₀ = Σ(n=1 to 20) (-1)ⁿ/((n-1)!) + Σ(n=1 to 20) (-1)ⁿ/n!</p><p>First sum (substitute m = n-1): Σ(m=0 to 19) (-1)^(m+1)/m! = -Σ(m=0 to 19) (-1)ᵐ/m!</p><p><strong>Step 3:</strong> Combine and simplify.</p><p>S₂₀ = -Σ(m=0 to 19) (-1)ᵐ/m! + Σ(n=1 to 20) (-1)ⁿ/n!</p><p>= -[1 - 1 + 1/2! - 1/3! + ... + (-1)¹⁹/19!] + [-1 + 1/2! - 1/3! + ... + (-1)²⁰/20!]</p><p>The middle terms cancel, leaving:</p><p>S₂₀ = -1 + (-1)²⁰/20! = -1 + 1/20!</p><p>∴ Answer: B</p>
Correct Answer: B