Binomial Theorem
Binomial Convolution
Grade 11
Question:
<p>The expression \((\binom{10}{0})^2 - (\binom{10}{1})^2 + (\binom{10}{2})^2 - \ldots + (\binom{10}{8})^2 - (\binom{10}{9})^2 + (\binom{10}{10})^2\) equals:</p>
<p>(a) \(10!\)</p>
<p>(b) \((\binom{10}{5})^2\)</p>
<p>(c) \(-\binom{10}{5}\)</p>
<p>(d) \(\binom{10}{5}\)</p>
Step-by-Step Solution
Key Concept: Use the coefficient of x^10 in the expansion of (1+x)^10(1-x)^10 = (1-x^2)^10, where the alternating sum of squares of binomial coefficients emerges as a coefficient. Recognize that this sum equals the coefficient of x^10 in (1-x^2)^10.
<p><strong>Step 1:</strong> Recognize the structure. We need to find:
$$\sum_{k=0}^{10} (-1)^k \binom{10}{k}^2$$</p>
<p><strong>Step 2:</strong> Use the generating function identity. Consider:
$$(1+x)^{10}(1-x)^{10} = [(1+x)(1-x)]^{10} = (1-x^2)^{10}$$</p>
<p><strong>Step 3:</strong> Expand the left side using binomial theorem:
$$(1+x)^{10} = \sum_{k=0}^{10} \binom{10}{k}x^k$$
$$(1-x)^{10} = \sum_{j=0}^{10} \binom{10}{j}(-x)^j = \sum_{j=0}^{10} (-1)^j\binom{10}{j}x^j$$</p>
<p><strong>Step 4:</strong> The coefficient of $x^{10}$ in $(1+x)^{10}(1-x)^{10}$ is:
$$\sum_{k=0}^{10} \binom{10}{k} \cdot (-1)^{10-k}\binom{10}{10-k} = \sum_{k=0}^{10} (-1)^{10-k}\binom{10}{k}^2$$</p>
<p><strong>Step 5:</strong> Since $(-1)^{10-k} = (-1)^{-k} \cdot (-1)^{10} = (-1)^k$ (as $(-1)^{10}=1$), we have:
$$\sum_{k=0}^{10} (-1)^k\binom{10}{k}^2$$
This is exactly our required sum.</p>
<p><strong>Step 6:</strong> Now expand the right side $(1-x^2)^{10}$:
$$(1-x^2)^{10} = \sum_{m=0}^{10} \binom{10}{m}(-x^2)^m = \sum_{m=0}^{10} (-1)^m\binom{10}{m}x^{2m}$$</p>
<p><strong>Step 7:</strong> The coefficient of $x^{10}$ in $(1-x^2)^{10}$ occurs when $2m=10$, so $m=5$:
$$\text{Coefficient of } x^{10} = (-1)^5\binom{10}{5} = -\binom{10}{5}$$</p>
<p><strong>Step 8:</strong> Therefore:
$$\sum_{k=0}^{10} (-1)^k\binom{10}{k}^2 = -\binom{10}{5}$$</p>
<p><strong>∴ Answer: C</strong></p>
Correct Answer: C