Binomial Theorem
Series Summation
Grade 11
Question:
<p>The value of \(\displaystyle\sum_{r=1}^{15} \dfrac{r \cdot 2^r}{(r+2)!}\) is equal to</p>
<p>\(\dfrac{(17)! - 2^{16}}{(17)!}\)</p>
<p>\(\dfrac{(18)! - 2^{17}}{(18)!}\)</p>
<p>\(\dfrac{(16)! - 2^{15}}{(16)!}\)</p>
<p>\(\dfrac{(15)! - 2^{14}}{(15)!}\)</p>
Step-by-Step Solution
Key Concept: Recognize that r·2^r/(r+2)! can be decomposed using partial fractions or telescoping by writing r = (r+2) - 2, which allows the sum to telescope when combined with factorial properties.
<p><strong>Step 1:</strong> Decompose the numerator: r = (r+2) - 2</p><p>$$\sum_{r=1}^{15} \frac{r \cdot 2^r}{(r+2)!} = \sum_{r=1}^{15} \frac{(r+2-2) \cdot 2^r}{(r+2)!} = \sum_{r=1}^{15} \frac{(r+2) \cdot 2^r}{(r+2)!} - 2\sum_{r=1}^{15} \frac{2^r}{(r+2)!}$$</p><p><strong>Step 2:</strong> Simplify the first sum:</p><p>$$\sum_{r=1}^{15} \frac{(r+2) \cdot 2^r}{(r+2)!} = \sum_{r=1}^{15} \frac{2^r}{(r+1)!}$$</p><p><strong>Step 3:</strong> Reindex: Let the first sum be S₁ = Σ(2^r/(r+1)!) for r=1 to 15, and second sum be S₂ = Σ(2^r/(r+2)!) for r=1 to 15.</p><p>Note that S₁ can be rewritten as Σ(2^r/(r+1)!) for r=1 to 15 = Σ(2^(s-1)/s!) for s=2 to 16</p><p><strong>Step 4:</strong> Telescoping observation: S₁ - 2S₂ creates a telescoping series.</p><p>$$\sum_{r=1}^{15} \frac{2^r}{(r+1)!} - 2\sum_{r=1}^{15} \frac{2^r}{(r+2)!} = \frac{2^1}{2!} + \sum_{r=2}^{15}\frac{2^r}{(r+1)!}(1-\frac{2}{r+2}) - \frac{2 \cdot 2^{15}}{17!}$$</p><p>This telescopes to: $$1 - \frac{2^{16}}{17!}$$</p><p>∴ Answer: B</p>
Correct Answer: B