Prof Gould combinatorial identity 3.27 and its "cousin" formula In the book on Combinatorial Identities of Prof Gould I found the identity 3.27
$$\sum_{k=0}^{\rho}\binom{2x+1}{2k+1}\binom{x-k}{\rho-k}=\frac{2x+1}{2\rho+1}\binom{x+\rho}{2\rho}2^{2\rho}$$ 
I now cant handle its "cousin" formula
$$\sum_{k=0}^{\rho}\binom{2x+1}{2k}\binom{x-k}{\rho-k}=?$$ and am looking for a similar identity. 
Maybe also the identity 3.26 is relevant or helpful in this context $$\sum_{k=0}^{\rho}\binom{2x}{2k}\binom{x-k}{\rho-k}=\frac{x}{x+\rho}\binom{x+\rho}{2\rho}2^{2\rho}$$ (Remark: to me looks this formula strangely different to 3.27 when I look at the denominator $x+\rho$)
 A: I think it's just:
$$
\sum_{k=0}^{\rho}\binom{2x+1}{2k}\binom{x-k}{\rho-k} = \binom{x+\rho}{2\rho}2^{2\rho}
$$
I did not try to prove it, but here is the reasoning: the term $ \binom{x+\rho}{2\rho}2^{2\rho} $ has to be there for asymptotic reasons. The fractional term in front I wasn't sure of, but I tried a few examples and it seems correct!
A: Suppose we seek to evaluate
$$Q(x,\rho) = \sum_{k=0}^\rho {2x+1\choose 2k}
{x-k\choose \rho-k}$$
where $x\ge\rho.$

Introduce
$${x-k\choose \rho-k} = {x-k\choose x-\rho} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{x-\rho+1}}
(1+z)^{x-k} \; dz.$$
Note that this controls the range being zero when $\rho\lt k \le x$ so
we can  extend the  sum to  $x$ supposing that  $x\gt \rho$.  And when
$x=\rho$ we may also set the upper limit to $x.$
We get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{x-\rho+1}}
(1+z)^{x} 
\sum_{k=0}^x {2x+1\choose 2k} \frac{1}{(1+z)^k}
\; dz.$$
This is
$$\frac{1}{2} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{x-\rho+1}}
(1+z)^{x} 
\left(\left(1+\frac{1}{\sqrt{1+z}}\right)^{2x+1}
+ \left(1-\frac{1}{\sqrt{1+z}}\right)^{2x+1}\right) \; dz
\\ = 
\frac{1}{2} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{x-\rho+1}}
\frac{1}{\sqrt{1+z}}
\left((1+\sqrt{1+z})^{2x+1}+(1-\sqrt{1+z})^{2x+1}\right) 
\; dz.$$
Observe that the second  term in the parenthesis (i.e. $1-\sqrt{1+z}$)
has no constant term and hence  starts at $z^{2x+1}$ making for a zero
contribution. This leaves
$$\frac{1}{2} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{x-\rho+1}}
\frac{1}{\sqrt{1+z}}
(1+\sqrt{1+z})^{2x+1}
\; dz.$$
Now put $1+z=w^2$ so that $dz = 2w\; dw$ to get
$$\frac{1}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(w^2-1)^{x-\rho+1}}
\frac{1}{w}
(1+w)^{2x+1}
\; w \; dw
\\ = \frac{1}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(w-1)^{x-\rho+1}}
\frac{1}{(w+1)^{x-\rho+1}}
(1+w)^{2x+1}
\; dw
\\ = \frac{1}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(w-1)^{x-\rho+1}}
(1+w)^{x+\rho} \; dw
\\ = \frac{1}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(w-1)^{x-\rho+1}}
\sum_{q=0}^{x+\rho}
{x+\rho\choose q} 2^{x+\rho-q} (w-1)^q 
\; dw.$$
This is
$$[(w-1)^{x-\rho}]
\sum_{q=0}^{x+\rho}
{x+\rho\choose q} 2^{x+\rho-q} (w-1)^q 
\\= {x+\rho\choose x-\rho} 2^{x+\rho-(x-\rho)}
= {x+\rho\choose x-\rho} 2^{2\rho}
= {x+\rho\choose 2\rho} 2^{2\rho}.$$
A: We can also prove the companion identity from above.

Suppose we seek to evaluate
$$Q(x,\rho) = \sum_{k=0}^\rho {2x+1\choose 2k+1}
{x-k\choose \rho-k}$$
where $x\ge\rho.$

Introduce
$${x-k\choose \rho-k} = {x-k\choose x-\rho} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{x-\rho+1}}
(1+z)^{x-k} \; dz.$$
Note that this controls the range being zero when $\rho\lt k \le x$ so
we can  extend the  sum to  $x$ supposing that  $x\gt \rho$.  And when
$x=\rho$ we may also set the upper limit to $x.$
We get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{x-\rho+1}}
(1+z)^{x} 
\sum_{k=0}^x {2x+1\choose 2k+1} \frac{1}{(1+z)^k}
\; dz.$$
This is
$$\frac{1}{2} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{x}}{z^{x-\rho+1}} \sqrt{1+z}
\left(\left(1+\frac{1}{\sqrt{1+z}}\right)^{2x+1}
- \left(1-\frac{1}{\sqrt{1+z}}\right)^{2x+1}\right) \; dz
\\ = 
\frac{1}{2} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{x-\rho+1}}
\left((1+\sqrt{1+z})^{2x+1}-(1-\sqrt{1+z})^{2x+1}\right) 
\; dz.$$
Observe that the second  term in the parenthesis (i.e. $1-\sqrt{1+z}$)
has no constant term and hence  starts at $z^{2x+1}$ making for a zero
contribution. This leaves
$$\frac{1}{2} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{x-\rho+1}}
(1+\sqrt{1+z})^{2x+1}
\; dz.$$
Now put $1+z=w^2$ so that $dz = 2w\; dw$ to get
$$\frac{1}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(w^2-1)^{x-\rho+1}}
(1+w)^{2x+1}
\; w \; dw
\\ = \frac{1}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(w-1)^{x-\rho+1}}
\frac{1}{(w+1)^{x-\rho+1}}
(1+w)^{2x+1}
\; w \; dw
\\ = \frac{1}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(w-1)^{x-\rho+1}}
(1+w)^{x+\rho} \; w \; dw.$$
Writing $w=(w-1)+1$ this produces two pieces, the first is
$$\frac{1}{2\pi i}
\int_{|w-1|=\epsilon}
\frac{1}{(w-1)^{x-\rho}}
\sum_{q=0}^{x+\rho}
{x+\rho\choose q} 2^{x+\rho-q} (w-1)^q 
\; dw.$$
This is
$$[(w-1)^{x-\rho-1}]
\sum_{q=0}^{x+\rho}
{x+\rho\choose q} 2^{x+\rho-q} (w-1)^q 
\\= {x+\rho\choose x-\rho-1} 2^{x+\rho-(x-\rho-1)}
= {x+\rho\choose x-\rho-1} 2^{2\rho+1}
= {x+\rho\choose 2\rho+1} 2^{2\rho+1}.$$
The second piece is
$$[(w-1)^{x-\rho}]
\sum_{q=0}^{x+\rho}
{x+\rho\choose q} 2^{x+\rho-q} (w-1)^q 
\\= {x+\rho\choose x-\rho} 2^{x+\rho-(x-\rho)}
= {x+\rho\choose x-\rho} 2^{2\rho}
= {x+\rho\choose 2\rho} 2^{2\rho}.$$
Joining the two pieces we finally obtain
$$\left(2\times \frac{x-\rho}{2\rho+1} + 1\right)
\times {x+\rho\choose 2\rho} 2^{2\rho}
\\ = \frac{2x+1}{2\rho +1}
{x+\rho\choose 2\rho} 2^{2\rho}.$$
