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 $(1)?$

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


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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.