<p>Find the coefficient of $x^{n-2}$ in the expansion of $f(x) = (x-1)(x-2)(x-3)\cdots(x-n)$</p>
Step-by-Step Solution
Key Concept: The coefficient of $x^{n-2}$ in the expansion of a product is the elementary symmetric polynomial $S_2$.
<p><strong>Step 1:</strong> Write $f(x) = x^n - S_1x^{n-1} + S_2x^{n-2} - \cdots + (-1)^n(1 \cdot 2 \cdot 3 \cdots n)$</p><p><strong>Step 2:</strong> where $S_1, S_2, \ldots$ are elementary symmetric polynomials in $1, 2, 3, \ldots, n$.</p><p><strong>Step 3:</strong> The coefficient of $x^{n-2}$ is $S_2$ = sum of products of first $n$ natural numbers taken 2 at a time</p><p><strong>Step 4:</strong> $S_2 = \frac{1}{2}[(1+2+\cdots+n)^2 - (1^2 + 2^2 + \cdots + n^2)]$</p><p><strong>Step 5:</strong> $= \frac{1}{2}\left[\left(\frac{n(n+1)}{2}\right)^2 - \frac{n(n+1)(2n+1)}{6}\right]$</p>
Correct Answer: S_2