Answer
a) $S_{n} = D(e^{-at} +e^{-2at} +...+e^{-nat})$
b) $S_{n} = \frac{D}{e^{at}-1}$
c) $D \ge C(e^{at} -1)$
Work Step by Step
a)
Let $S_{n}$ denote the residual concentration just before $(n+1)$-st injection.
$S_{1} = De^{-at}$
$S_{2} = De^{-at} + De^{-2at}$
$S_{3} = De^{-at} + De^{-2at} + De^{-3at}$
In general
$S_{n} = D(e^{-at} +e^{-2at} +...+e^{-nat})$
b)
$S_{n} = D(e^{-at} +e^{-2at} +...+e^{-nat})$
$a=De^{-at}$ and $r=e^{-at}$
$nā\infty$
$S_{n}=\frac{De^{-at}}{1-e^{-at}}=\frac{D}{e^{at}-1}$
c)
$\frac{D}{e^{at}-1} \ge C$
$D \ge C(e^{at} -1)$
