I was trying to derive the equation for not continuous compound interest. I started as follows: $$ P+P* \frac{R}{100x} = P[1+ \frac{R}{100x} ]$$ So $$ A = P[1+ \frac{R}{100x} ] + P[1+ \frac{R}{100x} ]* \frac{R}{100x} +P[1+ \frac{R}{100x} ]* [\frac{R}{100x}]^2+P[1+ \frac{R}{100x}]*[\frac{R}{100x}]^3+P[1+ \frac{R}{100x}]* [\frac{R}{100x}]^4+…+P[1+ \frac{R}{100x}]* [\frac{R}{100x}]^{x-1} $$ $$=P+2\frac{PR}{100x}+2[\frac{PR}{100x} ]^2+2[\frac{PR}{100x} ]^3+[\frac{PR}{100x}]^4+…+[\frac{PR}{100x} ]^x$$ Please note that the second $[\frac{PR}{100x}]^4$ is inside the $…$. I was not sure of if I should have still included it because it would appear with distribution of a term that was inside the $…$
Continuing the simplification
$$ A = P(1+2[\frac{R}{100x}]+2 [\frac{R}{100x}] ^2+2 [\frac{R}{100x}] ^3+ [\frac{R}{100x}]^4 +…+[\frac{R}{100x}]^x)$$
$$=P(\frac{R^x+…+R^4(100x)^{x-4}+2R^3(100x)^{x-3}+2R^2(100x)^{x-2}+2R(100x)^{x-1}+(100x)^x}{(100x)^x})$$
I do not understand how to simplify this further, however, as to get the normal equation the numerator would have to simplify to $(R+100x)^x$, but then how is it possible for all the coefficients to be 2 for the in between terms (the end terms have coefficient 1 as expected).
Go step-by-step. $$P\left(1+\frac{R}{100x}\right)+ P\left(1+\frac{R}{100x}\right) \frac{R}{100x}= P\left(1+\frac{R}{100x}\right) \left(1+\frac{R}{100x}\right)= P\left(1+\frac{R}{100x}\right)^2.$$
After this, the interest will be on $\displaystyle P\left(1+\frac{R}{100x}\right)^2$ and not on $\displaystyle P\left(1+\frac{R}{100x}\right) \frac{R}{100x}$.
So $$P\left(1+\frac{R}{100x}\right)^2+ P\left(1+\frac{R}{100x}\right)^2 \frac{R}{100x}= P\left(1+\frac{R}{100x}\right)^2 \left(1+\frac{R}{100x}\right)= P\left(1+\frac{R}{100x}\right)^3.$$
I hope you can take it from here.