Question:

<p><strong>Step 1: Decompose the sum</strong></p><p>We need to evaluate $$\sum_{r=0}^{20} r(20-r)\binom{20}{r}^2$$</p><p>Use the identities: $r\binom{20}{r} = 20\binom{19}{r-1}$ and $(20-r)\binom{20}{r} = 20\binom{19}{r}$</p><p><strong>Step 2: Apply identities</strong></p><p>$$r(20-r)\binom{20}{r}^2 = r\binom{20}{r} \cdot (20-r)\binom{20}{r} = 20\binom{19}{r-1} \cdot 20\binom{19}{r}$$</p><p>$$= 400\binom{19}{r-1}\binom{19}{r}$$</p><p><strong>Step 3: Sum over r</strong></p><p>$$\sum_{r=0}^{20} r(20-r)\binom{20}{r}^2 = 400\sum_{r=0}^{20}\binom{19}{r-1}\binom{19}{r}$$</p><p>Note: $\binom{19}{r-1} = 0$ when $r=0$, so the sum effectively runs from $r=1$ to $r=19$.</p><p><strong>Step 4: Use Vandermonde's identity variant</strong></p><p>By the Chu-Vandermonde convolution identity applied to products of binomial coefficients:</p><p>$$\sum_{r=1}^{19}\binom{19}{r-1}\binom{19}{r} = \text{coefficient of } x^{19} \text{ in } (1+x)^{19}(1+x)^{19}$$</p><p>This equals the coefficient of $x^{19}$ in $(1+x)^{38}$, which requires careful index adjustment.</p><p><strong>Step 5: Direct coefficient extraction</strong></p><p>$$\sum_{r=1}^{19}\binom{19}{r-1}\binom{19}{r} = \binom{39}{20}$$</p><p>This follows from the identity: $\sum_{k=0}^{n}\binom{n}{k}\binom{n}{k+1} = \binom{2n+1}{n+1}$ with $n=19$.</p><p><strong>Step 6: Final calculation</strong></p><p>$$\sum_{r=0}^{20} r(20-r)\binom{20}{r}^2 = 400 \cdot \binom{39}{20}$$</p><p><strong>∴ Answer:</strong> A</p>