Hello, and sorry for the delay.
These (F/A, i%, n) notations are awful.
I used to feel that way, until I programmed an application that calculated things for me easily (and that notation is very programming friendly). :)
Formulas (2.1) and (2.3) do not use the same notations as formula (3.3).
(2.1):
(F/P, i%, n) = F/P = (1 + i)^n
(2.3):
(F/A, i%, n) = F/A = [(1 + i)^n – 1]/i
And furthermore, formulas (2.1) and (2.3) should not be the basis of the derivation!
I had just listed them, with the goal in mind that whoever helped me knew what not to use.
You do not provide the exact meaning of (F/A, i%, n).
Therefore, it is impossible to answer your question without guessing what you exactly want.
In addition, this exercise seems to depend on the context of this book and I do not have a copy of it.
I suggest you to go back to the definitions of the (F/A, i%, n) and (F/P, i%, n).
by the way, what are these F, A, P?
I gave the algebraic definitions above. As for the theoretical definitions and meaning of variables from my book (Schaum’s Outline of Engineering Economics), they are as follows (with minor modifications).:
(F/P, i%, n):
Suppose that a given sum of money, P, earns interest at a rate i, compounded annually. The total amount of money, F (which I think is referred to as the “future worth”), which will have accumulated from an investment of P dollars after n years is given by F = P(1 + i)^n. The ratio F/P = (1 + i)^n is called the single-payment, compound-amount factor. The fuller notation for this ratio is (F/P, i%, n).
(F/A, i%, n):
Let equal amounts of money, A, be deposited in a savings account (or placed in some other interest-bearing investment) at the end of each year. If the money earns interest at a rate i, compounded annually, how much money will have accumulated after n years? To answer this question, we note that after n years, the first year’s deposit will have increased in value to F_1 = A(1 + i)^(n – 1). Similarly, the second year’s deposit will have increased in value to F_2 = A(1 + i)^(n – 2) and so on. The total amount accumulated will thus be the sum of a geometric progression.: F = F_1 + F_2 + . . . + F_n = A(1 + i)^(n – 1) + A(1 + i)^(n – 2) + . . . + A = A[(1 + i)^n – 1]/i. This implies that F/A = [(1 + i)^n – 1]/i. The ratio F/A = [(1 + i)^n – 1]/i is called the uniform-series, compound-amount factor. The extended notation for this ratio is (F/A, i%, n).
If I haven’t already done so, I can provide you with whatever else it is you need to know.