Hmm (expressing hesitation, etc.) ... This question seems a bit difficult, so I'll answer it even if it's homework. (I always feel that there seems to be a detailed theoretical analysis in the book of mathematical modeling, but the book is not at hand ...) First of all, this should be a mathematical experiment problem, not for you to seek an analytical solution, just for the problem of 20 years, and the last remaining money and the original money are satisfied:

Yu Qian of10 4 and Nest [# (105/100; , Jin Ben, 20];

Let's enter a random number and try:

yuqian[ 10^5.]

(* 1.3595* 10^ 13? *)

Far greater than 0. It is not difficult to analyze that the more money you save at the beginning, the more money you will have left after 20 years, that is, this function is monotonous, so FindRoot:

Find root[ Yu Qian [Jin Ben] == 0, {10 5}, Jin Ben, working accuracy-> 16]

(* {Jin Ben-> 16054.628 13746790}? *)

You can get the result without adding the precision option, that is, there will be a warning.

Then ask the second question. Since it's a math experiment, let's do it.

Yu Qian of10 4 and Nest [# (105/100; , Jin Ben,100];

Sol = find root[ Yu Qian [Jin Ben] == 0, {10 5}, Jin Ben, working accuracy-> 16,? max iterations-& gt； Infinite]

(* {Jin Ben-> 16055.02 148399739}*)

Note that the last option of FindRoot is required here.

Then we definitely want to know whether the result of 100 is valid in the future, then:

Yuqian list[ Jin Ben _]: = Nestlist [# (105/100.)-10 4&; , Jin Ben, 200];

List plot[Yuqian list[]/. Sol]

Well, the money has increased infinitely. In short, it is correct that the convergence radius is around 16055.

There will be a little difference according to the order of withdrawal and interest settlement. If it doesn't meet your requirements, you should know how to change it.