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?
Best Answer
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 $$ b_n=\sum_{k=0}^n \binom{n}{k} B_k (2^{n-k} -1 ) = n .$$