Binomial Theorem
Coefficient comparison
Grade 11
Question:
<p>If \(\displaystyle\sum_{r=0}^{n}\{a_r(x-\alpha+2)^r - b_r(\alpha-x-1)^r\} = 0\), then \(b_n\) is</p>
<p>(1) \(b_n = 1 + a_n\)</p>
<p>(2) \(b_n = (-1)^n \times a_n\)</p>
<p>(3) \(b_n = (-1)^{n-1} \times a_n\)</p>
<p>(4) \(b_n + 1 = a_n\)</p>
Step-by-Step Solution
Key Concept: Since the sum equals zero for all values of x, the coefficients of every power of x must individually equal zero. This means the two polynomials must be identical term-by-term, allowing us to match coefficients and find relationships between a_r and b_r.
<p><strong>Step 1:</strong> Recognize the relationship between the two expressions. Note that (α-x-1) = -(x-α+2).</p><p><strong>Step 2:</strong> Rewrite the second term: (-1)^r(α-x-1)^r = (-1)^r·(-(x-α+2))^r = (-1)^r·(-1)^r(x-α+2)^r = (x-α+2)^r</p><p><strong>Step 3:</strong> Substitute into the given equation: ∑_{r=0}^{n} {a_r(x-α+2)^r - b_r(x-α+2)^r} = 0</p><p><strong>Step 4:</strong> Factor: ∑_{r=0}^{n} (a_r - b_r)(x-α+2)^r = 0</p><p><strong>Step 5:</strong> For this to be zero for all x, we need a_r - b_r = 0 for all r. Therefore a_r = b_r for each r.</p><p><strong>Step 6:</strong> In particular, a_n = b_n. Since the given condition holds and the coefficient of (x-α+2)^n from the first sum is a_n and from the second is b_n, we have b_n = a_n.</p><p>∴ Answer: B</p>
Correct Answer: B