Sum with Bernoulli poly. $\sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n+1}{2k} B_{2k}(1/4) $ I have a roundabout proof for
$$ \sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n+1}{2k} B_{2k}(1/2)= \frac{n+1}{2^n} $$
where $B_{2k}(x)$ are the Bernoulli polynomials,$B_0(x)=1, B_2(x)=x^2-x+1/6,$ etc.
Changing the argument of the Bernoulli polynomial and observing patterns (aided by
OEIS A167205) I have a relationship which has
been checked to high values of positive integer $n,$
$$ \sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n+1}{2k} B_{2k}(1/4)= \frac{n+1}{2^{2n \ - \ (1 \ - \ (-1)^n)/2)}} \  \frac{3^n+1}{3- \ (-1)^n }$$
How can this equation be proved?

We seek to show that with $B_{2k}(x)$ a Bernoulli polynomial we have
$$\sum_{k=0}^{\lfloor n/2 \rfloor}
{n+1\choose 2k} B_{2k}(1/4) =
\frac{1}{2^{2n+1}} (n+1) (1+3^n).$$
We will use the EGF of the Bernoulli polynomials which is
$$\sum_{n\ge 0} B_n(x) \frac{z^n}{n!} =
\frac{z \exp(zx)}{\exp(z)-1}.$$
The LHS of the proposed identity is
$$(n+1) \sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose 2k} B_{2k}(1/4) \frac{1}{n+1-2k}$$
so we really only have to prove that
$$\sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose 2k} B_{2k}(1/4) \frac{1}{n+1-2k}
= \frac{1}{2^{2n+1}} (1+3^n).$$
We get two pieces for the LHS, the first call it A is
$$\frac{1}{2} \sum_{k=0}^{n}
{n\choose k} B_{k}(1/4) \frac{1}{n+1-k}$$
and the second call it B
$$\frac{1}{2} \sum_{k=0}^{n}
{n\choose k} (-1)^k B_{k}(1/4) \frac{1}{n+1-k}.$$
What we have here is a convolution of two exponential generating
functions. Recall that  when we  multiply two  exponential  generating
functions  of the  sequences $\{p_n\}$ and $\{q_n\}$ we get that
$$ P(z) Q(z) = \sum_{n\ge 0} p_n \frac{z^n}{n!}
\sum_{n\ge 0} q_n \frac{z^n}{n!}
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} p_k q_{n-k} z^n\\
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{n!}{k!(n-k)!} p_k q_{n-k} \frac{z^n}{n!}
= \sum_{n\ge 0}
\left(\sum_{k=0}^n {n\choose k} p_k q_{n-k}\right)\frac{z^n}{n!}$$
Observe that
$$\sum_{p\ge 0} \frac{1}{p+1} \frac{z^p}{p!} =
\frac{\exp(z)-1}{z}.$$
Therefore the EGF of piece A is
$$\frac{1}{2} \exp(z/4)$$
and the EGF of piece B is
$$\frac{1}{2} \frac{-z \exp(-z/4)}{\exp(-z)-1}
\frac{\exp(z)-1}{z}
= \frac{1}{2} \frac{-z \exp(3z/4)}{1-\exp(z)}
\frac{\exp(z)-1}{z}
\\ = \frac{1}{2} \exp(3z/4).$$
Extracting the coefficient on $[z^n]$ (EGF not OGF) on A and B
we find
$$\frac{1}{2} n! [z^n] (\exp(z/4) + \exp(3z/4)) =
\frac{1}{2^{2n+1}} (1+3^n)$$
as required.
