What is the accumulation factor (R) for an intermittent IV bolus with dose interval τ?

• A. R = 1 / (1 − e^{-kτ})
• B. R = e^{−kτ}
• C. R = 1 + e^{−kτ}
• D. R = 1 − e^{−kτ}

Answer: (A)

When a drug is given as intermittent IV boluses with first-order elimination, not all of the prior dose is cleared before the next dose. The amount remaining at the end of each dosing interval is multiplied by e^{-kτ}, so the contributions from all previous doses form a geometric series: 1 + e^{-kτ} + e^{-2kτ} + ... . The sum of this infinite series is 1 / (1 − e^{-kτ}), which defines the accumulation factor. This factor tells you how much higher the steady-state peak (or trough) is compared with a single-dose response.

Interpretation helps: if the dosing interval is long or the drug is cleared quickly, e^{-kτ} is small and R approaches 1 (little accumulation). If the interval is short or elimination is slow, e^{-kτ} is larger and R becomes much greater than 1 (significant accumulation).

Other forms don’t capture the cumulative effect of all prior doses. For example, e^{-kτ} alone is just the fraction remaining after one interval, not the summed contribution from all doses, and 1 − e^{-kτ} or 1 + e^{-kτ} don’t represent the infinite-sum accumulation.