Show that: $\sum_{l=0}^{m}(-1)^{m+l}4^{l-1}l{m \choose l}{2l \choose l}^{-1}{m+l \choose l}=\sum_{i=1}^{m}i^2$

Question

By observing and alteration of the OP's question we found this sum which is equivalent to the natural square number, $\sum_{i=1}^{m}i^2$

$$\sum_{l=0}^{m}(-1)^{m+l}4^{l-1}l{m \choose l}{2l \choose l}^{-1}{m+l \choose l}=\sum_{i=1}^{m}i^2\tag1$$

How can we prove $(1)?$

$$l{m \choose l}{2l \choose l}^{-1}{m+l \choose l}=\frac{m!}{l!(m-l)!}\cdot \frac{l!l!}{(2l)!}\cdot\frac{(m+l)!}{l!m!}=\frac{(m+l)!}{(m-l)!}\cdot\frac{l}{(2l)!}$$

$$\sum_{l=0}^{m}(-1)^{m+l}4^{l-1}\frac{(m+l)!}{(m-l)!}\cdot\frac{l}{(2l)!}\tag2$$

note that $$\frac{(m+l)!}{(m-l)!}\cdot\frac{l}{(2l)!}=\frac{(m+l)!(m+l+1)}{(m-l)!}\cdot\frac{l}{(2l)!(m+l+1)}=\frac{1}{B(m-l+1,2l+1)}\cdot \frac{l}{m+l+1}$$

$$\sum_{l=0}^{m}(-1)^{m+l}4^{l-1}\frac{1}{B(m-l+1,2l+1)}\cdot \frac{l}{m+l+1}\tag3$$

Where B(n,m) is the Beta function

Answer 1

We seek to evaluate

$$\sum_{q=0}^m (-1)^{m+q} 4^{q-1} q {m\choose q} {2q\choose q}^{-1} {m+q\choose q}.$$

We get from the binomial coefficients

$$\frac{m!}{(m-q)! \times q!} \frac{q! \times q!}{(2q)!} \frac{(m+q)!}{m! \times q!} = \frac{(m+q)!}{(m-q)! \times (2q)!} = {m+q\choose m-q}.$$

Our sum becomes

$$\sum_{q=0}^m (-1)^{m+q} 4^{q-1} q {m+q\choose m-q} = [z^m] (1+z)^m \sum_{q=0}^m (-1)^{m+q} 4^{q-1} q z^q (1+z)^q.$$

The coefficient extractor enforces the range and we find

$$[z^m] (1+z)^m \sum_{q\ge 0} (-1)^{m+q} 4^{q-1} q z^q (1+z)^q \\ = \frac{1}{4} (-1)^{m+1} [z^m] (1+z)^m \frac{4z(1+z)}{(1+4z(1+z))^2} \\ = (-1)^{m+1} [z^{m-1}] (1+z)^{m+1} \frac{1}{(1+2z)^4}.$$

This is

$$(-1)^{m+1} \mathrm{Res}_{z=0} \frac{1}{z^m} (1+z)^{m+1} \frac{1}{(1+2z)^4}.$$

Now we put $z/(1+z) = w$ so that $z = w/(1-w)$ and $dz = 1/(1-w)^2 \; dw$ to get

$$(-1)^{m+1} \mathrm{Res}_{w=0} \frac{1}{w^m} \frac{1}{1-w} \frac{1}{(1+2w/(1-w))^4}\frac{1}{(1-w)^2} \\ = (-1)^{m+1} \mathrm{Res}_{w=0} \frac{1}{w^m} \frac{1-w}{(1+w)^4} \\ = (-1)^{m+1} \left( [w^{m-1}] \frac{1}{(1+w)^4} - [w^{m-2}] \frac{1}{(1+w)^4} \right) \\ = [w^{m-1}] \frac{1}{(1-w)^4} + [w^{m-2}] \frac{1}{(1-w)^4} = {m+2\choose 3} + {m+1\choose 3} \\ = \frac{1}{6} m (m+1) (m+2+m-1) = \frac{1}{6} m (m+1) (2m+1) = \sum_{q=1}^m q^2.$$

This is the claim.