<p>If \((1-x^2)^n = \displaystyle\sum_{r=0}^{n} a_r x^r (1-x)^{2n-r}\), then \(a_r\) is equal to</p>
Step-by-Step Solution
Key Concept: Recognize that the given expansion requires expanding (1-x²)ⁿ in terms of binomial coefficients, then matching coefficients by comparing both sides after factoring out (1-x) terms strategically. The coefficient aᵣ depends on how many times the factor (1-x) appears in the right side.
<p><strong>Step 1:</strong> Start with (1-x²)ⁿ = ∑_{r=0}^{n} aᵣxʳ(1-x)^(2n-r)</p><p><strong>Step 2:</strong> Expand left side: (1-x²)ⁿ = ∑_{k=0}^{n} C(n,k)(-x²)ᵏ = ∑_{k=0}^{n} C(n,k)(-1)ᵏx^(2k)</p><p><strong>Step 3:</strong> For the right side, note that xʳ(1-x)^(2n-r) contains powers of x from degree r to degree r+(2n-r)=2n</p><p><strong>Step 4:</strong> To find aᵣ, we need the coefficient of x^(2r) from ∑aᵣx^r(1-x)^(2n-r). When r is even (say r=2m), the term a₂ₘx^(2m)(1-x)^(2n-2m) contributes to x^(2m) with coefficient a₂ₘ·C(2n-2m,0)=a₂ₘ</p><p><strong>Step 5:</strong> Comparing coefficients of x^(2m) on both sides: a₂ₘ = C(n,m)(-1)ᵐ</p><p><strong>Step 6:</strong> More generally, aᵣ = C(n, r/2)(-1)^(r/2) when r is even, and aᵣ = 0 when r is odd</p><p><strong>Step 7:</strong> This simplifies to: <strong>aᵣ = C(n, ⌊r/2⌋)(-1)^⌊r/2⌋</strong> or equivalently <strong>aᵣ = (-1)^(r/2)·C(n, r/2)</strong> for applicable values</p><p>∴ Answer: D</p>
Correct Answer: D