Binomial Theorem
Grade 11
Question:
<p>If C<sub>r</sub> stands for <sup>n</sup>C<sub>r</sub>, then the sum of the series <span class="math-tex">\(\frac{2\left(\frac{n}{2}\right) !\left(\frac{n}{2}\right) !}{n !}\left[C_{0}^{2}-2 C_{1}^{2}+3 C_{2}^{2}-\ldots+(-1)^{n}(n+1) C_{n}^{2}\right]\)</span>, where n is an even positive integer, is equal to</p>
<p style="display:inline">(-1)<sup>n</sup>(n + 1)</p>
<p style="display:inline">(-1)<sup>n/2</sup>(n + 1)</p>
<p style="display:inline">(-1)<sup>n/2</sup>(n + 2)</p>
<p style="display:inline">(1)<sup>n/2</sup>(n - 1)</p>
Step-by-Step Solution
Key Concept: Split the general term $(k+1)C_k^2$ into $C_k^2$ and $kC_k^2$ to leverage the alternating sum of squares identity and its derivative-based variant for even $n$.
<p>We have,<br />
<span class="math-tex">\(C_{0}^{2}-2 C_{1}^{2}+3 C_{2}^{2}-4 C_{3}^{2}+\ldots+(-1)^{n}(n+1) C_{n}^{2}\)</span><br />
<span class="math-tex">\(=\left[C_{0}^{2}-C_{1}^{2}+C_{2}^{2}-C_{3}^{2}+\ldots+(-1)^{n} C_{n}^{2}\right]\)</span> <span class="math-tex">\(-\left[C_{1}^{2}-2 C_{2}^{2}+3 C_{3}^{2}-\ldots+(-1)^{n} n C_{n}^{2}\right]\)</span><br />
<span class="math-tex">\(=(-1)^{n / 2} \frac{n !}{\left(\frac{n}{2}\right) !\left(\frac{n}{2}\right) !}-(-1)^{\frac{n}{2}-1} \frac{n}{2} \frac{n !}{\left(\frac{n}{2}\right) !\left(\frac{n}{2}\right) !}\)</span><br />
<span class="math-tex">\(=(-1)^{n/ 2} \frac{n !}{\left(\frac{n}{2}\right) !\left(\frac{n}{2}\right) !}\left(1+\frac{n}{2}\right)\)</span><br />
<span class="math-tex">\(\therefore \frac{2\left(\frac{n}{2}\right) !\left(\frac{n}{2}\right) !}{n !}\left[C_{0}^{2}-2 C_{1}^{2}+3 C_{2}^{2}-\ldots+(-1)^{r}(n+1) C_{n}^{2}\right]\)</span><br />
<span class="math-tex">\(=\frac{2\left(\frac{n}{2}\right) !\left(\frac{n}{2}\right) !}{n !}(-1)^{n / 2} \frac{n !}{\left(\frac{n}{2}\right) !\left(\frac{n}{2}\right) !} \frac{(n+2)}{2}=(-1)^{n / 2}(n+2)\)</span></p>
Correct Answer: C