Calculus 8th Edition

Amount of drugs inside the body after $3$ tablets is $157.875$mg. and Amount of drugs inside the body after $n$ tablets is $A(n) =\frac{3000}{19}[1-0.05^{n}]$
$A(n) = 150 \times [1+0.05^{2} + 0.05^{3} +...+0.05^{n-1}]$ $A(n) = 150 \times [\Sigma^{n}_{i=1} 0.05^{i-1}]$ $A(n) = 150 \times [\frac{1-0.05^{n}}{1-0.05}]$ $A(n) = 150 \times [\frac{1-0.05^{n}}{0.95}]$ $A(n) = \frac{3000}{19}[1-0.05^{n}]$ Now, $A(3) = \frac{3000}{19}[1-0.05^{3}] \approx 157.875$mg Hence, the amount of drugs inside the body after $3$ tablets is $157.875$mg and the amount of drugs inside the body after $n$ tablets is $A(n) =\frac{3000}{19}[1-0.05^{n}]$.