To determine the value of \( s \) in the expression \( \sum_{k=1}^{n} k(k+1)(k-1) = pn^4 + qn^3 + tn^2 + sn \), we first simplify the summation. We note that \( k(k+1)(k-1) = k(k^2-1) = k^3-k \). Consequently, the summation becomes: \[ \sum_{k=1}^{n} k(k+1)(k-1) = \sum_{k=1}^{n} (k^3-k) = \sum_{k=1}^{n} k^3 - \sum_{k=1}^{n} k. \] Employing the standard formulas: \[ \sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2 \quad \text{and} \quad \sum_{k=1}^{n} k = \frac{n(n+1)}{2}, \] we substitute these into the expression: \[ \left( \frac{n(n+1)}{2} \right)^2 - \frac{n(n+1)}{2}. \] Combining these terms by factoring out \(\frac{n(n+1)}{2}\): \[ \frac{n(n+1)}{2} \left( \frac{n(n+1)}{2} - 1 \right). \] Expanding the terms within the parentheses yields: \[ \frac{n(n+1)}{2} \left( \frac{n^2+n - 2}{2} \right) = \frac{n(n+1)(n^2+n-2)}{4}. \] Further expansion results in: \[ = \frac{n(n^3+n^2-2n+n^2+n-2)}{4} = \frac{n(n^3+2n^2-n-2)}{4}. \] Expanding completely: \[ = \frac{n^4 + 2n^3 - n^2 - 2n}{4}. \] Comparing this to the polynomial \( pn^4 + qn^3 + tn^2 + sn \), the coefficient of \( n \), which is \( s \), is \(-2/4\), simplifying to \(-1/2\).