Permutations & Combinations
Partial fractions and combinatorial identities
Grade 11
Question:
<p>In the identity \(\displaystyle\sum_{k=0}^{n} \frac{A_k}{x+k} = \frac{n!}{x(x+1)(x+2)\cdots(x+n)}\), the value of \(A_i\) is</p>
<p>(a) \({}^nC_i\)</p>
<p>(b) \({}^nC_{i+1}\)</p>
<p>(c) \((-1)^i \cdot {}^nC_i\)</p>
<p>(d) \((-1)^{i-1} \cdot {}^nC_{i-1}\)</p>
Step-by-Step Solution
Key Concept: Use partial fraction decomposition by multiplying both sides by (x)(x+1)...(x+n) and comparing coefficients, or recognize that A_i is the residue at the pole x = -i in the partial fraction expansion.
<p><strong>Step 1:</strong> Multiply both sides by the denominator x(x+1)(x+2)⋯(x+n):</p><p>∑_{k=0}^{n} A_k · [x(x+1)⋯(x+n)]/(x+k) = n!</p><p><strong>Step 2:</strong> The left side becomes: ∑_{k=0}^{n} A_k · x(x+1)⋯(x+k-1)(x+k+1)⋯(x+n)</p><p><strong>Step 3:</strong> To find A_i, substitute x = -i:</p><p>A_i · (-i)(-i+1)⋯(-i+i-1)(-i+i+1)⋯(-i+n) = n!</p><p>A_i · (-i)(-i+1)⋯(-1)(1)(2)⋯(n-i) = n!</p><p><strong>Step 4:</strong> Simplify the product:</p><p>A_i · [(-1)^i · i!] · [(n-i)!] = n!</p><p>A_i = n! / [(-1)^i · i! · (n-i)!] = (-1)^i · C(n,i)</p><p><strong>Step 5:</strong> Therefore: A_i = (-1)^i · inom{n}{i}</p><p>∴ Answer: A_i = (-1)^i inom{n}{i}</p>
Correct Answer: C