Let \(n \ge 2\) be an integer and \(p_n (x)\) be the polynomial \[p_n (x) = (x-1) + (x-2) + \cdots + (x-n).\] What is the remainder when \(p_n (x)\) is divided by \(p_{n-1} (x)\)?

\(\dfrac{n}{2}\);

\(\dfrac{n+1}{2}\);

\(\dfrac{n^2+n}{2}\);

\(\dfrac{-n}{2}\).

We can use the formula for the sum of an arithmetic progression to calculate \[p_n(x) = nx - \frac{n(n+1)}{2}\] and \[p_{n-1}(x) = (n-1)x - \frac{n(n-1)}{2}.\]

#### Approach 1

We can carry out long division, by noting that the \(x\) term of \(p_{n-1}(x)\) divides \(\dfrac{n}{n-1}\) times into the \(x\) term of \(p_n(x)\), so that the remainder is \[\begin{align*} p_n(x) - \dfrac{n}{n-1}p_{n-1}(x) &{}= \left(nx - \dfrac{n(n+1)}{2}\right) - \dfrac{n}{n-1} \left((n-1)x - \dfrac{n(n-1)}{2}\right)\\ &{}= -\dfrac{n(n+1)}{2} + \dfrac{n^2}{2}\\ &{}= \dfrac{-n}{2}, \end{align*}\]and the answer is (d).

#### Approach 2

We can use the Remainder Theorem. We have that \[p_{n-1}(x)=(n-1)\bigl(x-\tfrac{1}{2}n\bigr),\] and so \(p_{n-1}(x) =0\) when \(x = \dfrac{n}{2}\) (we can ignore the \(n-1\) factor as it will not change the remainder).

So the remainder is equal to \(p_n\left(\dfrac{n}{2}\right) = -\dfrac{n}{2}\), as before.

The answer is (d).