(a)

#### To determine

**To find:** The sum of the residual concentrations just before the (n+1)st injection.

#### Answer

The sum of the residual concentrations just before the (n+1)st injection is DeaT(1−e−anT)1−e−aT.

#### Explanation

**Given:**

The concentration of insulin in a patient’s system is De−at.

**Result used:**

The sum of the finite geometric series (∑k=1nark−1 (or) a+ar+ar2+⋯+arn−1) is a(1−rn)1−r, where *a* is the first term and *r* is the common ratio of the series.

**Calculation:**

The residual concentration just before the second injection is De−aT.

The residual concentration just before the third injection is De−a2T.

The residual concentration just before the fourth injection is De−a3T.

Proceed in the similar way, the residual concentrations just before the (n+1)st injection is De−anT.

The sum of the residual concentrations just before the (n+1)st injection Cn is computed as follows,

Cn=De−aT+De−a2T+De−a3T+⋯+De−a(n−1)T+De−anT=D(e−aT)1+D(e−aT)2+D(e−aT)3+⋯+D(e−aT)n−1+D(e−aT)n=∑k=1nD(e−aT)k

Clearly, it is finite geometric series with the first term is a=De−aT and the common ratio is r=e−aT.

Use the Result stated above and obtain the sum of the series ∑k=1nD(e−aT)k.

∑k=1nD(e−aT)k=De−aT(1−(e−aT)n)1−e−aT=De−aT(1−e−anT)1−e−aT

Therefore, Cn=De−aT(1−e−anT)1−e−aT.

Hence, the sum of the residual concentrations just before the (n+1)st injection is DeaT(1−e−anT)1−e−aT.

(b)

#### To determine

The limiting pre-injection concentration.

#### Answer

The limiting pre-injection concentration is De−aT1−e−aT.

#### Explanation

From part (a), the sum of the residual concentrations just before the (n+1)st injection is

Cn=De−aT(1−e−anT)1−e−aT

Obtain the limit of the term Cn as *n* tends to infinity.

limn→∞Cn=limn→∞De−aT(1−e−anT)1−e−aT=De−aT(1−e−∞)1−e−aT=De−aT(1−0)1−e−aT=De−aT1−e−aT

Divide the numerator and the denominator by e−aT,

limn→∞Cn=De−aTe−aT1−e−aTe−aT=D1e−aT−e−aTe−aT=DeaT−1

Therefore, the limiting pre-injection concentration is DeaT−1.

(c)

#### To determine

The minimal dosage *D* in terms of *C*, *a* and *T.*

#### Answer

The minimal dosage is D=C(eaT−1).

#### Explanation

Obtain the minimal dosage *D*.

From part (b), the concentration of insulin is DeaT−1.

Note that, DeaT−1 must always remain at or above the critical value *C.* That is, DeaT−1≥C.

Multiply the inequality by (eaT−1),

DeaT−1⋅(eaT−1)≥C⋅(eaT−1)D≥C(eaT−1)

Therefore, the minimal dosage is D=C(eaT−1).