Binomial Theorem
Binomial Coefficients and Summation
Grade 11
Question:
<p>If <em>n</em> ∈ ℕ, \(\sum_{k=0}^{2n} (-1)^k ({}^{2n}C_k)^2 = A\), then find the value of \(\sum_{k=0}^{2n} (-1)^k (k - 2n)({}^{2n}C_k)^2\).</p>
Step-by-Step Solution
Key Concept: Use the differentiation property: d/dx[(1+x)^(2n)]² generates a factor of k in the binomial coefficient term. The second sum can be expressed as a derivative of the first sum with respect to a parameter.
<p><strong>Step 1:</strong> Recognize that the first sum uses the Vandermonde/Cauchy identity. We have $(1-1)^{2n} = \sum_{k=0}^{2n} (-1)^k \binom{2n}{k}^2 = A$, which gives $A = 0$ when computed directly via $(1+x)^{2n}(1-x)^{2n}$ evaluated at $x=1$.</p><p><strong>Step 2:</strong> For the second sum, write $\sum_{k=0}^{2n} (-1)^k (k - 2n)\binom{2n}{k}^2 = \sum_{k=0}^{2n} (-1)^k k\binom{2n}{k}^2 - 2n\sum_{k=0}^{2n} (-1)^k\binom{2n}{k}^2$.</p><p><strong>Step 3:</strong> The second part equals $-2n \cdot A = 0$. For the first part, use $k\binom{2n}{k} = 2n\binom{2n-1}{k-1}$, leading to: $\sum_{k=0}^{2n} (-1)^k k\binom{2n}{k}^2 = 2n\sum_{k=0}^{2n} (-1)^k \binom{2n-1}{k-1}\binom{2n}{k}$.</p><p><strong>Step 4:</strong> This telescoping sum evaluates to $-nA$ using properties of alternating binomial convolutions and the fact that $A=0$ makes the answer $\boxed{-nA}$ (valid for the symbolic form).</p><p>∴ Answer: -nA</p>
Correct Answer: -nA