<p>The sum \(2 \times {}^{40}C_2 + 6 \times {}^{40}C_3 + 12 \times {}^{40}C_4 + 20 \times {}^{40}C_5 + \ldots + 1560 \times {}^{40}C_{40}\) is divisible by</p>
Step-by-Step Solution
Key Concept: Recognize that the coefficients 2, 6, 12, 20, ... follow the pattern k(k+1) for k = 1, 2, 3, ... and use the identity k(k+1)·C(n,k) = n(n-1)·C(n-2,k-2) to convert this to a binomial expansion.
<p><strong>Step 1:</strong> Identify the coefficient pattern. The coefficients are 2=1·2, 6=2·3, 12=3·4, 20=4·5, ..., 1560=39·40. These are of the form k(k+1) for k = 1, 2, 3, ..., 39.</p><p><strong>Step 2:</strong> Rewrite the sum as S = Σ k(k+1)·C(40,k) for k=1 to 39 (note: C(40,1)=40, and 1·2·40=80, which appears to be the first term).</p><p><strong>Step 3:</strong> Use the identity k(k+1)·C(n,k) = n(n-1)·C(n-2,k-2). Here, k(k+1)·C(40,k) = 40·39·C(38,k-2).</p><p><strong>Step 4:</strong> Therefore S = 40·39·Σ C(38,k-2) for k=1 to 39, which equals 40·39·Σ C(38,j) for j=-1 to 37. The sum of all binomial coefficients C(38,j) for j=0 to 38 is 2^38, but we need j from -1 to 37, giving us 40·39·(2^38 - C(38,38)) = 40·39·(2^38 - 1).</p><p><strong>Step 5:</strong> S = 40·39·(2^38-1) = 1560·(2^38-1). This is divisible by 1560, 40, 39, 2^38-1, and their factors including 5, 8, 13, 78, 195, 390, etc.</p><p>∴ The sum is divisible by factors of 1560 including: 2, 4, 5, 8, 10, 13, 20, 26, 39, 40, 52, 65, 78, 130, 156, 195, 260, 390, 520, 780, 1560</p>
Correct Answer: A,B,C,D