How to prove the equivalence of the two functions? $$f_1(k,n):=\sum_{0\leqslant v\leqslant n}\dfrac{\left(2(k+v)\right)!}{(k+v)!v!(2k+v)!(n-v)!2^v}, \quad k,n \in \mathbb{N}
$$
$$
f_2(k,n):=\sum_{0\leqslant m\leqslant \lfloor{\frac{n}{2}}\rfloor}\dfrac{1}{(k+m)!m!(n-2m)!2^{4m-n}},\quad k,n \in \mathbb{N}
$$
I checked many small natural number (up to several hundred) $k$'s and $n$'s, and all indicate the two functions are equivalent.
How to prove or disprove it?
 A: Suppose we seek to verify that $f_1(n,k) = f_2(n,k)$ 
where
$$f_1(n,k) = \sum_{v=0}^n 
\frac{(2k+2v)!} {(k+v)!\times v!\times (2k+v)!\times (n-v)!} 2^{-v}$$
and
$$f_2(n,k) = \sum_{m=0}^{\lfloor n/2\rfloor} 
\frac{1}{(k+m)!\times m! \times (n-2m)!} 2^{n-4m}.$$
Multiplying by $(n+k)!$ we obtain
$$g_1(n,k) = \sum_{v=0}^n {n+k\choose n-v} {2k+2v\choose v} 2^{-v}$$
and
$$g_2(n,k) = 2^n \sum_{m=0}^{\lfloor n/2\rfloor} 
{n+k\choose m} {n+k-m\choose n-2m} 2^{-4m}.$$
We will work with the latter two.

Re-write the first sum as follows:
$$2^{-n} \sum_{v=0}^n {n+k\choose v} 
{2k+2n-2v\choose n-v} 2^{v}$$
Introduce
$${2k+2n-2v\choose n-v} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-v+1}} (1+z)^{2k+2n-2v} \; dz.$$
This integral is zero when $v\gt n$ so we may extend $v$ to infinity.
We get for $g_1(n,k)$
$$2^{-n} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} (1+z)^{2k+2n} 
\sum_{v\ge 0} {n+k\choose v} \frac{z^v}{(1+z)^{2v}} 2^v
\; dz
\\ = 2^{-n} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} (1+z)^{2k+2n} 
\left(1+2\frac{z}{(1+z)^2}\right)^{n+k}
\; dz
\\ = 2^{-n} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
\left(1+4z+z^2\right)^{n+k}
\; dz.$$
For the second sum introduce
$${n+k-m\choose n-2m} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-2m+1}} (1+z)^{n+k-m} \; dz.$$
This is zero when $2m\gt n$ so we may extend $m$ to infinity.
We get for $g_2(n,k)$
$$2^n \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} (1+z)^{n+k} 
\sum_{m\ge 0} {n+k\choose m} \frac{z^{2m}}{(1+z)^m} 2^{-4m} 
\; dz
\\ = 2^n \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} (1+z)^{n+k} 
\left(1+\frac{1}{16}\frac{z^2}{1+z}\right)^{n+k}
\; dz
\\ = 2^n \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
\left(1+z+\frac{1}{16}z^2\right)^{n+k}
\; dz.$$
Finally put $z=4w$ in this integral to get
$$2^n \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{4^{n+1} w^{n+1}}
\left(1+4w+w^2\right)^{n+k}
\; 4 dw
\\ = 2^{-n} \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n+1}}
\left(1+4w+w^2\right)^{n+k}
\; dw.$$
This concludes the argument.
