$\displaystyle \sum_{r=0}^{n}\binom{2n}{2r}(2-4^r)\beta_{2r}=2^{2n}\beta_{2n} \\$
I am trying to prove this sum. Actually It's a conjecture made by me, And I'm pretty sure it works. Now I tried to evaluate the sum but I hardly find any way out. Any help is really appreciated .
Also the related sum $\displaystyle \sum_{r=0}^{n}\binom{2n+1}{2r}(2-4^r)\beta_{2r}=0$ is already provided but without any proof. I guess the proofs are inter related. I will make an edit if I get something
We introduce the generating function
$$G(z) = \sum_{n\ge 0} \frac{z^{2n}}{(2n)!} \sum_{r=0}^{n} {2n\choose 2r} (2-2^{2r}) B_{2r}.$$
This is
$$\sum_{n\ge 0} z^{2n} \sum_{r=0}^{n} \frac{1}{(2n-2r)!} (2-2^{2r}) \frac{1}{(2r)!} B_{2r} \\ = \sum_{n\ge 0} z^{2n} \sum_{r=0}^{n} \frac{1}{(2n-2r)!} (2-2^{2r}) [w^{2r}] \frac{w}{\exp(w)-1} \\ = \sum_{r\ge 0} (2-2^{2r}) [w^{2r}] \frac{w}{\exp(w)-1} \sum_{n\ge r} \frac{z^{2n}}{(2n-2r)!} \\ = \sum_{r\ge 0} (2-2^{2r}) z^{2r} [w^{2r}] \frac{w}{\exp(w)-1} \sum_{n\ge 0} \frac{z^{2n}}{(2n)!} \\ = \frac{1}{2} (\exp(z)+\exp(-z)) \sum_{r\ge 0} (2-2^{2r}) z^{2r} [w^{2r}] \left(\frac{1}{2} w + \frac{w}{\exp(w)-1}\right) \\ = \frac{1}{2} (\exp(z)+\exp(-z)) \left(z + \frac{2z}{\exp(z)-1} - z - \frac{2z}{\exp(2z)-1} \right) \\ = z (\exp(z)+\exp(-z)) \left(\frac{1}{\exp(z)-1} - \frac{1}{\exp(2z)-1} \right) \\ = z (\exp(z)+\exp(-z)) \frac{\exp(z)}{\exp(2z)-1} = z \frac{\exp(2z)+1}{\exp(2z)-1} \\ = z \left(1+\frac{2}{\exp(2z)-1}\right).$$
It follows that
$$\bbox[5px,border:2px solid #00A000]{ (2n)! [z^{2n}] G(z) = (2n)! [z^{2n}] \frac{2z}{\exp(2z)-1}}$$
which is
$$\bbox[5px,border:2px solid #00A000]{ 2^{2n} (2n)! [z^{2n}] \frac{z}{\exp(z)-1} = 2^{2n} B_{2n}}$$
and this is the claim.