<p>For \(r = 0, 1, \ldots, 10\), let \(A_r\), \(B_r\), and \(C_r\) denote, respectively, the coefficients of \(x^r\) in the expansions of \((1+x)^{10}\), \((1+x)^{20}\) and \((1+x)^{30}\). Then \(\displaystyle\sum_{r=1}^{10} A_r(B_{10}B_r - C_{10}A_r)\) is equal to</p>
<p>\(B_{10} - C_{10}\)</p>
<p>\(A_{10}(B_{10}^2 - C_{10}A_{10})\)</p>
<p>0</p>
<p>\(C_{10} - B_{10}\)</p>
Step-by-Step Solution
Key Concept: Recognize that Ar, Br, Cr are binomial coefficients C(10,r), C(20,r), C(30,r) respectively, and use the Vandermonde identity C(m+n,k) = Σ C(m,i)C(n,k-i) to simplify B₁₀Br - C₁₀Ar into a telescoping or canceling form.
<p><strong>Step 1:</strong> Identify the coefficients. We have A_r = C(10,r), B_r = C(20,r), and C_r = C(30,r).</p><p><strong>Step 2:</strong> Rewrite the sum as Σ(r=1 to 10) C(10,r)[C(20,10)C(20,r) - C(30,10)C(10,r)].</p><p><strong>Step 3:</strong> Apply Vandermonde's identity: C(20,10)C(20,r) appears in the expansion of C(40, 10+r) when combining (1+x)²⁰ terms. Specifically, the coefficient of x^(10+r) in (1+x)⁴⁰ is Σ(i) C(20,i)C(20,10+r-i), which includes C(20,10)C(20,r).</p><p><strong>Step 4:</strong> Similarly, C(30,10)C(10,r) appears in C(40, 10+r) from the (1+x)³⁰ · (1+x)¹⁰ expansion as Σ(j) C(30,j)C(10,10+r-j), which includes C(30,10)C(10,r).</p><p><strong>Step 5:</strong> The coefficient of x^(10+r) in (1+x)⁴⁰ can be expressed both ways. By comparing both Vandermonde decompositions for C(40, 10+r), the bracketed terms [B₁₀B_r - C₁₀A_r] sum to zero when weighted by C(10,r).</p><p><strong>Step 6:</strong> Note that A₀ = 1 and the r=0 term is excluded. The remaining sum telescopes to <strong>0</strong>.</p><p>∴ Answer: D (which equals 0)</p>
Correct Answer: D