Binomial Theorem
Sum of binomial coefficients
Grade 11
Question:
<p>If \(C_r = {}^nC_r\), the sum of the series \(\dfrac{2\left[\left(\dfrac{n}{2}\right)!\right]^2}{n!}\left[C_0^2 - 2C_1^2 + 3C_2^2 - \ldots + (-1)^n(n+1)C_n^2\right]\), where \(n\) is an even integer, is</p>
<p>(a) 0</p>
<p>(b) \((-1)^{n/2} \cdot (n+1)\)</p>
<p>(c) \((-1)^n \cdot (n+2)\)</p>
<p>(d) \((-1)^n \cdot n\)</p>
Step-by-Step Solution
Key Concept: Recognize that the alternating sum involving squared binomial coefficients can be extracted from the coefficient of x^n in the expansion of (1-x)^n(1+x)^n = (1-x²)^n, combined with differentiation to introduce the (r+1) factor.
<p><strong>Step 1:</strong> Recognize the series structure. We need: $\sum_{r=0}^{n}(-1)^r(r+1)C_r^2$</p><p><strong>Step 2:</strong> Use the identity $(1+x)^n(1-x)^n = (1-x^2)^n$. The coefficient of $x^n$ in $(1+x)^n(1-x)^n$ gives $\sum_{r=0}^{n}(-1)^rC_r^2$.</p><p><strong>Step 3:</strong> To get the $(r+1)$ factor, differentiate. Consider $\frac{d}{dx}[x(1-x^2)^n] = (1-x^2)^n + x \cdot n(-2x)(1-x^2)^{n-1}$. Evaluate the coefficient of $x^{n-1}$ (which corresponds to $x^n$ after multiplying by $x$).</p><p><strong>Step 4:</strong> For even $n$, the coefficient of $x^n$ in $(1-x^2)^n$ is $(-1)^{n/2}\binom{n}{n/2}$.</p><p><strong>Step 5:</strong> Therefore: $\sum_{r=0}^{n}(-1)^r(r+1)C_r^2 = (-1)^{n/2}\binom{n}{n/2}$</p><p><strong>Step 6:</strong> Substitute into the original expression: $\frac{2[(n/2)!]^2}{n!} \cdot (-1)^{n/2}\binom{n}{n/2} = \frac{2[(n/2)!]^2}{n!} \cdot (-1)^{n/2} \cdot \frac{n!}{(n/2)!(n/2)!}$</p><p><strong>Step 7:</strong> Simplify: $= 2(-1)^{n/2}$</p><p>∴ Answer: <strong>B</strong></p>
Correct Answer: B