Consider the following pseudo code. What is the total number of multiplications to be performed?D = 2 for i = 1 to n do for j = i to n do for k = j + 1 to n do D = D * 3

To find the total number of multiplications, count the number of times the statement D = D * 3 is executed.

The statement executes for each combination of indices $(i, j, k)$ such that:

For a fixed $i$ and $j$, $k$ runs from $j+1$ to $n$, which gives $(n - j)$ executions.

Summing over $j$ from $i$ to $n$:

Now summing over $i$ from $1$ to $n$: Let $p = n - i$. As $i$ varies from $1$ to $n$, $p$ takes values from $n - 1$ down to $0$:

Using the identity $\sum_{p=1}^{m} p(p+1) = \frac{m(m+1)(m+2)}{3}$ with $m = n - 1$:

Here, $(n-1)n(n+1)$ is the product of three consecutive integers. Thus, the total number of multiplications is one-sixth of the product of the 3 consecutive integers.

Hence, the correct option is (C).