Direct formula to calculate total amount for compound interest with a compounding interest rate

Question

Principal amount: $100
Compound interest rate: 24%
Compound interest on the original interest: 2%


Total amount after first year: 100 * 1.24 = $124
Total amount after second year: 100 * 1.24 * (1 + (0.24*1.02)) = $154.3552 (Because interest rate also grows)
Total amount after third year: 100 * 1.24 * (1+(0.24*1.02)) * (1+(0.24*1.02*1.02)) = $192.89707

Is it possible to derive a direct formula to calculate the total accumulated amount after $n$ number of years? A formula without iterative steps (summation etc), one where I can directly feed in $n$ and get the answer using a basic calculator.

The formula that I tried was $P \times(1.24^n) \times (1.02^{n(n-1) / 2})$, this is wrong for obvious reasons, one can not take out 1.02 factor directly outside, it adds extra interest for original principal rather than just to the interest rate. Is it possible to achieve a direct formula for this?

Note: I am new to the math stack exchange, kindly adjust with the math conventions that I might not have followed

Answer 1

I believe the 'formula' is..

$$ P * \prod_{n=1}^{n} (1 + (0.24){(1 + 0.02)}^{n-1} )$$

As Dave Says, I do not think much can be done to simplify it to create a formula without using the product notation. The 1 + parts in the rate, make it hard to simplify it, unlike when principal compounds by number and also that number compounds by a number.

Your formula in the question

$$ P \times(1.24^n) \times (1.02^{n(n-1) / 2}) $$

will work when

P is principal amount
1.24 is the base factor by which the amount compounds every month
1.02 is the factor by which the base factor compounds every month

Edit 1:

Assuming $0.24$ as $r$, and $1.02$ as $q$ I tried the product series in wolfram to see if it offers a simplification. It offered a new notation called $q-Pochhammer$, which I am not familiar with. But looking around about it, I do not think even that offers a simplification to a formula that looks like the one in the question.