<p>The value of \(\displaystyle\sum_{r=0}^{10}(-1)^r \cdot 4^{10-r} \cdot {}^{30}C_r \cdot {}^{30}C_{10-r}\) is equal to</p>
<p>(1) \({}^{30}C_{10} \times 2^{10}\)</p>
<p>(2) \({}^{30}C_9 \times 4^{10}\)</p>
<p>(3) \({}^{30}C_{10} \times 3^{10}\)</p>
<p>(4) \({}^{30}C_9 \times 3^{10}\)</p>
Step-by-Step Solution
Key Concept: Recognize this sum as the coefficient of x^10 in the expansion of (1-4)^30 using the convolution property of binomial coefficients. The alternating signs with decreasing powers of 4 and the product of binomial coefficients indicate a Cauchy product structure.
<p><strong>Step 1:</strong> Recognize the structure. We have ∑_{r=0}^{10}(-1)^r · 4^{10-r} · C(30,r) · C(30,10-r).</p><p><strong>Step 2:</strong> Rewrite as ∑_{r=0}^{10}C(30,r) · C(30,10-r) · (-1)^r · 4^{10-r} = 4^{10}∑_{r=0}^{10}C(30,r) · C(30,10-r) · (-1/4)^r.</p><p><strong>Step 3:</strong> Apply Vandermonde's convolution: ∑_{r=0}^{10}C(30,r) · C(30,10-r) is the coefficient of x^{10} in (1+x)^{30}·(1+x)^{30} = (1+x)^{60}.</p><p><strong>Step 4:</strong> This equals C(60,10). But we need ∑_{r=0}^{10}C(30,r) · C(30,10-r) · (-1/4)^r, which is the coefficient of x^{10} in (1+x)^{30}·(1-x/4)^{30} = [(1+x)(1-x/4)]^{30}·(1+x)^0 evaluated properly.</p><p><strong>Step 5:</strong> Reframe: coefficient of x^{10} in (1+x)^{30}·(1-x/4)^{30}. Let u = 1+x, then (1-x/4) = 1-(u-1)/4 = (3+u)/4. We seek [u(3+u)/4]^{30}, coefficient of u^{10}.</p><p><strong>Step 6:</strong> Direct approach: coefficient of t^{10} in (1+t)^{30}(1-t/4)^{30} = coefficient of t^{10} in [(4-t)/4]^{30} = (1/4^{30})·(-1)^{10}·C(30,10)·4^{20} = C(30,10).</p><p>∴ Answer: <strong>C(30,10)</strong></p>
Correct Answer: C