I hope to find the sum of a series defined by $(1+a)^n$. So far, using binomial expansion the function expands quite nicely into: $$(1+a)^n=\binom{n}{0}a^0+\binom{n}{n}a^n+\binom{n}{1}a+...+\binom{n}{n-1}a^{n-1}$$ $$(1+a)^n=1+a^n+\frac{n!}{1!(n-1)!}\cdot(a+a^{n-1})+\frac{n!}{2!(n-2)!}\cdot(a^2+a^{n-2})+...+k$$ where $k$ is some general term which largely depends on whether n is odd or even. I can't quite write it down yet.
I was wondering if you could find the sum for the series, like as you would with: $$\sum_{i=0}^n \frac{1}{(1+a)^i} = \frac{1-(1+a)^{-n}}{a}$$
Not sure if I make any sense.
$$(1+a)^1=\binom 10a^0+\binom 11a^1=\binom 10(a^0+a^1),\\ (1+a)^2=\binom 20a^0+\binom 21a^1+\binom 22a^2=\binom 20(a^0+a^2)+\binom 21a^1.$$
For an odd exponent, there is no "$k$" term and there are $(n+1)/2$ pairs with equal coefficients; and for an even exponent, the $k$ term is $$\binom n{n/2}a^{n/2}$$ plus $n/2$ pairs.
Anyway, in all cases the sum is just expressed by $(1+a)^n$.