Prof Gould combinatorial identity 3.27 and its "cousin" formula

Question

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 $x+\rho$)

Answer 1

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!

Answer 2

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 $$[(w-1)^{x-\rho}] \sum_{q=0}^{x+\rho} {x+\rho\choose q} 2^{x+\rho-q} (w-1)^q \\= {x+\rho\choose x-\rho} 2^{x+\rho-(x-\rho)} = {x+\rho\choose x-\rho} 2^{2\rho} = {x+\rho\choose 2\rho} 2^{2\rho}.$$

Answer 3

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 $$\left(2\times \frac{x-\rho}{2\rho+1} + 1\right) \times {x+\rho\choose 2\rho} 2^{2\rho} \\ = \frac{2x+1}{2\rho +1} {x+\rho\choose 2\rho} 2^{2\rho}.$$