Solution Verification: Show that $\sum_{i=0}^{n}{\binom{n+a}{n-i}\binom{i+a+1}{i}}=2^{n}\binom{n+a}{a}+2^{n-1}\binom{n+a}{a+1}$ for $n\geq 1$ I solved it by counting the same object using different methods. Say that there are $n+a$ boys and $1$ girl. I want to form a team of $a+1$ kids with some reserve players, but I want the girl to be either in the team or at least be a reserve player.
Method 1
I first select $i+a$ boys, then from these and the girl ($i+a+1$ kids) I select $a+1$ to be in the team while the rest ($i$ kids) will be the reserve players. The number of possibilities is;
$\sum_{i=0}^{n}{\binom{n+a}{i+a}\binom{i+a+1}{a+1}}= \sum_{i=0}^{n}{\binom{n+a}{n-i}\binom{i+a+1}{i}} $
Method 2
If the girl is in the team, then I need to select $a$ boys to complete the team while the remaining $n$ boys can be reserve players or not. The number of possibilities is;
$2^{n}\binom{n+a}{a}$
If the girl is only a reserve player, I need to select $a+1$ boys to be in the team while the remaining $n-1$ boys can be reserve players or not. The number of possibilities is;
$2^{n-1}\binom{n+a}{a+1}$
Conclusion
Since both method count the same objects then I can conclude that;
$\sum_{i=0}^{n}{\binom{n+a}{n-i}\binom{i+a+1}{i}}=2^{n}\binom{n+a}{a}+2^{n-1}\binom{n+a}{a+1}$
I want to know if my solution is correct and if there’s any different solution out there.
 A: OP asks for an alternate evaluation of
$$\sum_{q=0}^n {n+a\choose n-q} {a+1+q\choose q}.$$
We have by inspection that this is
$$[z^n] (1+z)^{n+a} \frac{1}{(1-z)^{a+2}}$$
which is in turn
$$\underset{z}{\mathrm{res}}\;
\frac{1}{z^{n+1}} (1+z)^{n+a}  \frac{1}{(1-z)^{a+2}}.$$
Now we put $z/(1+z) = w$ so that $z=w/(1-w)$ and $dz = 1/(1-w)^2 \; dw$
to get
$$\underset{w}{\mathrm{res}}\;
\frac{1}{w^n} \frac{1-w}{w}
\frac{1}{(1-w)^a} 
\frac{(1-w)^{a+2}}{(1-2w)^{a+2}}
\frac{1}{(1-w)^2}
\\ = \underset{w}{\mathrm{res}}\;
\frac{1-w}{w^{n+1}}
\frac{1}{(1-2w)^{a+2}}.$$
This is
$$[w^n] (1-w) \frac{1}{(1-2w)^{a+2}}
= 2^n {n+a+1\choose a+1} - 2^{n-1} {n+a\choose a+1}
\\ = 2^n \frac{n+a+1}{a+1} {n+a\choose a}
- 2^{n-1} {n+a\choose a+1}
\\ = 2^n {n+a\choose a}
+ 2^n \frac{n}{a+1} {n+a\choose a}
- 2^{n-1} {n+a\choose a+1}
\\ = 2^n {n+a\choose a} + 2^{n-1} {n+a\choose a+1}$$
which is the claim.
A: Inspired by Marko Riedel’s answer, I recognized that both expression give the coefficient of $x^{a+1}$ from  $\left(1+x\right)\left(2+x\right)^{n+a}$ so we can conclude that
$$
\sum_{i=0}^{n}{\binom{n+a}{n-i}\binom{i+a+1}{i}}=2^{n}\binom{n+a}{a}+2^{n-1}\binom{n+a}{a+1}
$$
In addition, we can use the same method to get another result:
$$
\sum_{i=0}^{n}{\left(-1\right)^{i}\binom{n+a}{n-i}\binom{i+a+1}{i}}=0\phantom{x}\phantom{x}\forall\phantom{x}n\geq2
$$
