I came up with an interesting question for myself but, I got stumbled when trying to solve it. I'm not advanced in maths so forgive me.
You are investing in something which gives you back a certain amount.
Let's represent this with $xr$
$x$ being the amount invested in and $r$ being the rate. However, for the first $t$ years you put that investment back into $x$ giving you a system such as the following..
$(x_0)r = x_1$
$\left(x_0+x_1\right) r = x_2$
$(x_0+x_1+x_2)r = x_3$
I tried to put it together into one big formula to calculate $t$ however, that variable of $r$ is giving me a headache. How would I come up with a formula for calculating the value at any given time or $t$ If my math isn't horribly wrong; if $t=5$ then the value would be $=xr+4xr^2+6xr^3+4xr^4+xr^5$
Is there a way to solve for $t = n$, where $n$ is a real number with a formula?
This is normal savings with accumulated interest, and quite well known:
$$ x_1 = x_0r\\ x_2 = (x_0+x_1)r\\ x_3 = (x_0+x_1+x_2)r $$
Instead of looking at how much the investment gives, try looking at how much money you have, let's call that $t_0, t_1, \ldots$, then you have $$ t_0 = x_0\\ t_1 = t_0+t_0r = t_0(1+r)\\ t_2 = t_1+t_1r = t_1(1+r) = t_0(1+r)(1+r) = t_0(1+r)^2 $$
From here it doesn't take long to conclude that in general $t_n=t_{n-1}(1+r)=t_0(1+r)^n$.
Using the binomial formula, we get: $$ (1+r)^n=\sum_{i=0}^{n} \binom{n}{i}\cdot 1^i\cdot\binom{n}{n-i}\cdot r^{n-i} $$ which is exactly the polynomials you've be getting.