Question:

<p><strong>Step 1: Recognize the pattern as a generating function coefficient</strong></p><p>The given sum represents the coefficient of x^r in:</p><p>$$\sum_{k=0}^{n} (-1)^k \binom{n}{k} x^{2k} \cdot (1+x)^{2n-2k}$$</p><p>This is equivalent to finding the coefficient of x^r in: $(1-x^2)^n(1+x)^{2n}$</p><p><strong>Step 2: Simplify the generating function</strong></p><p>$$(1-x^2)^n(1+x)^{2n} = (1-x)^n(1+x)^n(1+x)^{2n}$$</p><p>$$= [(1-x)(1+x)]^n(1+x)^{2n} = (1-x^2)^n(1+x)^{2n}$$</p><p><strong>Step 3: Expand (1-x²)^n</strong></p><p>$$(1-x^2)^n = \sum_{j=0}^{n} \binom{n}{j}(-1)^j x^{2j}$$</p><p>The expansion contains only even powers of x: 0, 2, 4, 6, ..., 2n.</p><p><strong>Step 4: Determine when coefficient of x^r is zero</strong></p><p>For the product $(1-x^2)^n(1+x)^{2n}$ to have a non-zero coefficient at x^r:</p><p>- We need x^r to come from x^{2j} (even power from first factor) times x^{r-2j} (from second factor)</p><p>- If r is odd, no combination of even power (2j) plus any power (r-2j) from $(1+x)^{2n}$ can yield x^r</p><p>- More precisely, (1-x^2)^n contributes only even powers, so multiplying by (1+x)^{2n} preserves the parity constraint</p><p><strong>Step 5: Analyze when r is odd and r < 2n</strong></p><p>When r is odd, say r = 2m+1, the coefficient of x^r in $(1-x^2)^n(1+x)^{2n}$ vanishes because:</p><p>- The first factor contributes only even powers</p><p>- No combination of even + any power = odd power in the product structure yields x^r for odd r</p><p><strong>Step 6: Verification with special case</strong></p><p>For r = 1 (odd): The sum equals $\binom{2n}{1} - \binom{n}{1}\binom{2n-2}{1} + ... = 0$</p><p>$$\therefore \text{Answer: D}$$</p>