The total value of all the payments and the accumulated interest, at $i=0.0025,$ is given by$$100+100(1+i)+100(1+i)^2+100(1+i)^3+...+100(1+i)^{n-1}$$The last payment accumulates zero interest, and the first payment compounds for $n-1$ months. The sum of this geometric series, $S$, is given by:$$S=\frac{100((1+i)^n-1).}{i}$$ You then need to subtract $100n$ from this total to find the interest portion.
While the solutions to approach this through "brute force" are fine for a small number of periods, I thought I'd show a more general approach when you're asked to do the computation for longer durations. If you find the following heavy-going, then just do the brute force computation (I indicate where you can immediately do this in the working below).
Approach this problem from first principles. An annual interest rate of $12\%$ compounded monthly is a compound monthly rate of $1\%$. A monthly compounding means the amount owed will grow by a factor of $(1 + 0.01) = 1.01$ with each month.
Let the initial amount at the beginning of the first month (the principal) be represented by $A_0$.
Then the amount at the end of the first month $A_1 = 1.01A_0 - 4200$.
The last term is because he does a partial repayment by $\$4200$.
In fact, this is the general formula for the recursion:
$A_{n+1} = 1.01A_n - 4200$.
At this point, you can immediately apply this recursive calculation to find $A_3$, which is the simple "brute force" approach. Use $A_0$ to find $A_1$, then $A_1$ to find $A_2$, then finally $A_2$ to find $A_3$.
But if you want a more general solution, you can work through this to find a closed form for $A_n$.
$A_n = 1.01A_{n-1} - 4200 = 1.01^2A_{n-2} - 4200(1.01 + 1) \\= 1.01^3A_{n-3} - 4200(1.01^2 + 1.01 + 1) \\= \dots \\= 1.01^nA_0 - 4200(1.01^{n-1} + 1.01^{n-2} + \dots + 1.01^0)$
the last term of which can be further simplified by the geometric sum formula.
$$4200(1.01^{n-1} + 1.01^{n-2} + \dots + 1.01^0) = 4200\frac{1.01^n-1}{1.01-1} = 4.2 \times 10^5(1.01^n-1)$$
After a little rearrangement, you will get:
$A_n = 4.2 \times 10^5 - 1.01^n(4.2 \times 10^5 - A_0)$
This closed form can be proved using induction for rigour.
From this, you can find out $A_3 = 4.2 \times 10^5 - 1.01^3(4.2 \times 10^5 - 50000) = \$38,788.63$. This is the answer to your question.
If you pursue this to the end (by setting $A_n = 0$), you will find he will have fully repaid his loan by the $13$th month.
Best Answer
If your bank deposits only yield $\ 4.8\%\ $ per year interest, then your goal of deriving an income stream therefrom which increases at a rate of $\ 20\%\ $ per year indefinitely is not achievable. Suppose the income you derive from interest in the first year is $\ \epsilon\ $, and let $\ n\ $ be the smallest positive integer such that $$ \left(\frac{150}{131}\right)^n> \frac{1,000,000}{\epsilon}\ . $$ Then after $\ n\ $ years, your income stream is required to be $\ 1.2^n\epsilon\ $ per annum, but even if all the interest obtained from the bank were devoted to growing the principal, then after $\ n\ $ years the principal would only be $\ 1.048^n\cdot1,000,000\ $, which is less than the income of $\ 1.2^n\epsilon\ $ it is required to generate (because $\ 1.2^n\epsilon = \left(\frac{150}{125}\right)^n\epsilon >$$ \left(\frac{131}{125}\right)^n\cdot1,000,000\ $$=1.048^n\cdot1,000,000\ $).