I have the following sum:
$$\sum_{i=0}^{n-1}\bigg\{\binom{n-1}{n-1-i}a^{(n-1-i)}\bigg[(1-a)^{i}\bigg(\frac{1-m^{n-i}}{n-i}\bigg)+\frac{m^{n-i}y^{i}}{n}\bigg]\bigg\}$$
It looks (obviously) quite similar to the binomial theorem in summation form. Any ideas on how to reduce this and express it as something other than a sum?
This isn't a full answer, but I can't type it in a comment. I'm not sure I'm good enough at MathJax to type it at all. Let $$\begin{align} A_i&=\binom{n-1}{n-1-i}a^{n-1-i}\\ B_i&=(1-a)^i\left(\frac{1-m^{n-i}}{n-i}\right)\\ C_i&=\frac{m^{n-i}y^i}{n} \end{align}$$ Then the given expression is $$ \sum_{i=0}^{n-1}{A_i(B_i+C_i)}=\sum_{i=0}^{n-1}A_iB_i+\sum_{i=0}^{n-1}A_iC_i$$
It was the second term in the last expression that I was addressing in my comment. $$\sum_{i=0}^{n-1}A_iC_i=\sum_{i=0}^{n-1}{\binom{n-1}{n-1-i}a^{n-1-i}}\frac{m^{n-i}y^i}{n}=\\ \frac{m}{n}\sum_{i=0}^{n-1}{\binom{n-1}{n-1-i}a^{n-1-i}m^{n-1-i}y^i}=\frac{m}{n}(am+y)^{n-1}, $$ by the binomial theorem. So the original expression simplifies to $$\sum_{i=0}^{n-1}A_iB_i+\frac{m}{n}(am+y)^{n-1}$$
I was trying to say that in my comment without typing all this.
EDIT I believe I know how to do the second part also. Let $$ f(x,y)=\sum_{k=0}^{n-1}{\binom{n-1}{k}\frac{x^{n-k}y^k}{n-k}},\text{ so that}\\ f_x(x,y)=\sum_{k=0}^{n-1}{\binom{n-1}{k}x^{n-1-k}y^k}=(x+y)^{n-1}\text{ and}\\ f(x,y)=\frac{(x+y)^n}{n}+C(y)\\ $$ Since f(0,y)=0, we have $C(y)=\frac{-y^n}{n}$ and $$ \boxed{f(x,y)=\frac{(x+y)^n-y^n}{n}}$$
Now you can write $\sum_{i=0}^{n-1}A_iB_i$ as the sum of two terms, each of which is expressible in the form $f(x,y)$ times some constant.