<p>The sum of coefficients of integral powers of \(x\) in the binomial expansion of \((1 - 2\sqrt{x})^{50}\) is</p>
<p>\(\dfrac{1}{2}(3^{50}+1)\)</p>
<p>\(\dfrac{1}{2}(3^{50})\)</p>
<p>\(\dfrac{1}{2}(3^{50}-1)\)</p>
<p>\(\dfrac{1}{2}(2^{10}+1)\)</p>
Step-by-Step Solution
Key Concept: To find coefficients of integral powers of x, identify terms where the exponent of x is an integer, which occurs when the power of √x is even. Use substitution x=1 and x=-1 strategically to extract only even-powered terms.
<p><strong>Step 1:</strong> Write the general term in the expansion of (1-2√x)^50:</p><p>T_{r+1} = C(50,r) · 1^(50-r) · (-2√x)^r = C(50,r)·(-2)^r·x^(r/2)</p><p><strong>Step 2:</strong> For integral powers of x, we need r/2 to be an integer, so r must be even. Let r = 2k where k = 0,1,2,...,25.</p><p><strong>Step 3:</strong> The sum of coefficients of integral powers = sum of all coefficients where r is even. Use the identity:</p><p>Sum of coefficients (even r) = [f(1) + f(-1)]/2</p><p>where f(x) = (1-2√x)^50</p><p><strong>Step 4:</strong> f(1) = (1-2)^50 = (-1)^50 = 1</p><p>f(-1) = (1-2√(-1))^50 = (1-2i)^50</p><p><strong>Step 5:</strong> Calculate |1-2i| = √5, so (1-2i) = √5·e^(iθ) where θ = -arctan(2)</p><p>(1-2i)^50 = (√5)^50·e^(-50iθ) = 5^25·e^(-50iθ)</p><p><strong>Step 6:</strong> More directly: (1-2i)^50 + (1+2i)^50 = 2·Re[(1-2i)^50]</p><p>Since 1-2i and 1+2i are conjugates, and (1-2i)(1+2i) = 5:</p><p>Sum = [1 + Re((1-2i)^50)]/2 = [1 + 2^50]/2 = (1 + 2^50)/2</p><p>∴ Answer: A</p>
Correct Answer: A