Sequences & Series
Summation using Binomial Coefficients
Grade 11
Question:
<p>If \(S_1 = \sum_{j=1}^{10} j(j-1)\dfrac{10!}{j(j-1)(j-2)!(10-j)!}\), \(S_2 = \sum_{j=1}^{10} j \dfrac{10!}{(j-1)!(9-(j-1))!}\), and \(S_3 = \sum_{j=1}^{10}[j(j-1)+j]\dfrac{10!}{j!(10-j)!}\), find \(S_3\).</p>
<p>\(55 \cdot 2^9\)</p>
<p>\(110 \cdot 2^8\)</p>
<p>\(55 \cdot 2^8\)</p>
<p>\(110 \cdot 2^9\)</p>
Step-by-Step Solution
Key Concept: Recognize that the factorials in each sum represent binomial coefficients; simplify each sum using binomial identities and the relationship between consecutive binomial terms, then combine results using the property that S₃ = S₁ + S₂.
<p><strong>Step 1: Simplify S₁</strong></p><p>For S₁, note that j(j-1) appears in both numerator and denominator:</p><p>S₁ = ∑(j=1 to 10) j(j-1) · 10!/[j(j-1)(j-2)!(10-j)!] = ∑(j=1 to 10) 10!/[(j-2)!(10-j)!]</p><p>This requires j ≥ 2. Rewrite: S₁ = ∑(j=2 to 10) 10·9·8!/[(j-2)!(10-j)!] = 90∑(k=0 to 8) C(8,k) = 90·2⁸ = 23040</p><p><strong>Step 2: Simplify S₂</strong></p><p>S₂ = ∑(j=1 to 10) j · 10!/[(j-1)!(9-(j-1))!] = ∑(j=1 to 10) j · 10!/[(j-1)!(10-j)!]</p><p>= ∑(j=1 to 10) 10·9!/[(j-1)!(10-j)!] = 10∑(j=1 to 10) C(9,j-1) = 10·2⁹ = 5120</p><p><strong>Step 3: Express S₃ in terms of S₁ and S₂</strong></p><p>S₃ = ∑(j=1 to 10) [j(j-1)+j] · 10!/[j!(10-j)!]</p><p>= ∑(j=1 to 10) j(j-1) · 10!/[j!(10-j)!] + ∑(j=1 to 10) j · 10!/[j!(10-j)!]</p><p>The first sum equals S₁ (after index adjustment) and second equals S₂ (after index adjustment)</p><p><strong>Step 4: Calculate final answer</strong></p><p>S₃ = S₁ + S₂ = 23040 + 5120 = <strong>28160</strong></p><p>∴ Answer: A</p>
Correct Answer: A