Binomial Theorem
Sum of Coefficients
Grade 11
Question:
<p>The sum of coefficients of integral powers of \(x\) in the binomial expansion of \((1-2\sqrt{x})^{50}\) is</p>
<p>\(\dfrac{1}{2}(3^{50})\)</p>
<p>\(\dfrac{1}{2}(3^{50}-1)\)</p>
<p>\(\dfrac{1}{2}(2^{50}+1)\)</p>
<p>\(\dfrac{1}{2}(3^{50}+1)\)</p>
Step-by-Step Solution
Key Concept: Separate integral and non-integral power terms by using x = 1 and x = -1 strategically. The term (1-2√x)^50 generates powers of √x, so we need to identify which terms yield integer powers and sum their coefficients.
<p><strong>Step 1:</strong> Substitute √x = t, so x = t². The expansion becomes (1-2t)^50 = Σ C(50,r)(-2t)^r = Σ C(50,r)(-2)^r·t^r</p><p><strong>Step 2:</strong> Integral powers of x = t² occur when r is even. So we need coefficients of even powers of t.</p><p><strong>Step 3:</strong> Let f(t) = (1-2t)^50 = Σ C(50,r)(-2)^r·t^r. To extract even-powered terms:</p><ul><li>f(1) = (1-2)^50 = (-1)^50 = 1 gives sum of all coefficients</li><li>f(-1) = (1+2)^50 = 3^50 gives alternating sum</li></ul><p><strong>Step 4:</strong> Sum of coefficients of even powers of t = [f(1) + f(-1)]/2 = (1 + 3^50)/2</p><p><strong>Step 5:</strong> These are exactly the coefficients of integral powers of x.</p><p>∴ Answer: <strong>(1 + 3^50)/2</strong></p>
Correct Answer: D