Binomial Transform of a Particular Binomial Sum Let
$$
a_n = \sum_{k=0}^n \binom{n + 1}{k}b_k.
$$
I am trying to write $b_n$ in terms of $a_k$.
Of course, if the binomial coefficient was $\binom{n}{k}$ instead of $\binom{n + 1}{k}$, then this is simple to do using the binomial transform,
$$
c_n = \sum_{k=0}^n\binom{n}{k}d_k \iff d_n = \sum_{k=0}^n\binom{n}{k}(-1)^{n-k}c_k.
$$
But in my case I'm at a loss. I tried using Pascal's identity, thinking I might be able to use the formula for the binomial transform if I could get $a_n$ expressed in terms of $b_k$ in the right way. This got me to
$$
a_n = \sum_{k=0}^n \binom{n}{k}b_k + \sum_{k=0}^{n-1}\binom{n}{k}b_{k + 1},
$$
which is almost right, but not quite. Is there a simple way to do this?
 A: What you seek is not quite a binomial transform.  Use the elementary ID
$$ \binom{n+1}{k} = \frac{n+1}{n+1-k} \binom{n}{k}.$$
Define $A_n=a_n/(n+1)$ and the sum you want to invert is
$$ A_n = \sum_{k=0}^n \binom{n}{k} \frac{b_k}{n-k+1} = \sum_{k=0}^n \binom{n}{k} \frac{b_{n-k}}{k+1} $$
where in the last step the sum has been performed in opposite order. Now Riordan, 'Combinatorial Identities,' eq 3.4.(17) states the inversion
$$ A_n = \sum_{k=0}^n \binom{n}{k} \frac{b_{n-k}}{k+1} \implies
 b_n = \sum_{k=0}^n \binom{n}{k} B_k A_{n-k} $$
where $B_n$ are the Bernoulli numbers $B_0=1, \ B_1=-1/2, \  B_{2n+1} = 0.$
Example: $a_n=(2^n-1)(n+1),$ $A_n=2^{n}-1,$ then
$$ b_n=\sum_{k=0}^n \binom{n}{k} B_k (2^{n-k} -1 ) = n .$$
A: Given
$$ a_n = \sum_{k=0}^n \binom{n + 1}{k}b_k. \tag{1} $$
Let the exponential generating functions (e.g.f.) for $\,a_n\,$ and $\,b_n\,$ be
$$ A(x) := \sum_{n=0}^\infty a_n x^n/n! \quad \text{ and }
  \quad B(x) := \sum_{n=0}^\infty b_n x^n/n!. \tag{2} $$
Multiply both sides of equation $(1)$ by $\,\dfrac{x^{n+1}}{(n+1)!}\,$
and sum over $\,n\ge0\,$ to get
$$ \int_0^x A(t)\,dt = (e^x-1) B(x). \tag{3} $$
Thus
$$ B(x) = \frac{x}{e^x-1} \left(\frac1x \int_0^x A(t)\,dt\right) \tag{4} $$
where $\,\dfrac{x}{e^x-1}\,$ is the e.g.f. of the Bernoulli numbers.
The 2nd factor is the e.g.f. of $\,\dfrac{a_n}{n+1}.\,$ Thus,
$$ b_n = \sum_{k=0}^n \binom{n}{k} B_{n-k} \frac{a_k}{k+1}. \tag{5} $$

Note: For the binomial transform, $\,A(x) = e^x B(x)\,$ is equivalent
to $\, B(x) = e^{-x}A(x).\,$
