Perhaps the easiest way to see what’s going on is to look at an extreme case. Suppose that you start with just two initial values, $10$ and $1000$, and increase one of them by $10$% while leaving the other unchanged. The average of the two percentage increases is $\frac12(10+0)=5$% no matter which of the two numbers you increased. The percentage increase in the total, however, very clearly depends on which of the two was increased: increasing the $10$ by $10$% increases the total by $1$, from $1010$ to $1011$, but increasing the $1000$ by $10$% increases the total by $100$, from $1010$ to $1110$. Clearly the latter is a much larger percentage increase than the former; the actual figures are about $0.099$% in the first case and about $9.9$% in the second.
The point is that when you increase one of the items by a certain percentage, the effect on the total depends not only on the percentage, but also on the size of the item.
Edit: And that’s exactly why your weighted averages work: the individual percentage increases need to be weighted by the sizes of the original items. Let the original amounts be $a_1,\dots,a_n$, and let the percentage increases, expressed as fractions, be $p_1,\dots,p_n$. The amounts after the increase are $a_k(1+p_k)=a_k+a_kp_k$ for $k=1,\dots,n$, so the the total increase is $$a_1p_1+a_2p_2+\cdots+a_np_n\;.$$ Let $A=a_1+a_2+\cdots+a_n$ be the original total; then the fractional increase is
$$\frac{a_1p_1+a_2p_2+\cdots+a_np_n}A=\frac{a_1}Ap_1+\cdots+\frac{a_n}Ap_n\;,$$
which is indeed the weighted mean of the percentages when they are weighted by the relative sizes of the original amounts.
Because dividing by $1$ doesn't do anything, so you might as well not do it. If the weights didn't add to $1$, you would have to divide by their sum, but in this special case you don't have to. If you really want to, you can, of course.
Best Answer
First of all you can put $\alpha$ and $(1-\alpha)^n$ in front of the sigma sign:
$$(1-\alpha)^n + (1-\alpha)^{n}\cdot \alpha\sum_{i=1}^n (1-\alpha)^{-i} $$
Factoring out $(1+\alpha)^n$
$$(1-\alpha)^n \left(1+\alpha\sum_{i=1}^n (1-\alpha)^{-i}\right) $$
That means, that the brackets must be $(1-\alpha)^{-n}$, since $(1-\alpha)^n\cdot (1-\alpha)^{-n}=1 \quad (\color{red}*)$
The sum is the partial sum of a $\color{grey}{\textrm{geometric series}}$, where $\sum\limits_{i=1}^n x^i=x\cdot \frac{x^n-1}{x-1}$. Thus
$$1+\alpha\sum_{i=1}^n \left(\frac1{1-\alpha}\right)^{i}=1+\frac{\alpha}{1-\alpha}\frac{\left(\frac1{1-\alpha}\right)^{n}-1}{\frac1{1-\alpha}-1}$$
$(1-a)\cdot \left( \frac1{1-\alpha}-1\right)=1-1+a=a$. Therefore the denominator is $\alpha$.
$$1+\alpha\frac{\left(\frac1{1-\alpha}\right)^{n}-1}{\alpha}=1+\left(\frac1{1-\alpha}\right)^{n}-1=\left(\frac1{1-\alpha}\right)^{n}=(1-\alpha)^{-n}$$
This is was we wanted at $(\color{red}*)$