Inverting a binomial sum Fix $n.$ Suppose I have two sequences of positive integers $a_k$ and $b_k$ that vanish for $k>n$ and satisfy the following relation: 
$$ a_k = \sum_{i=k}^n (-1)^{i-k} \binom{i}{k} b_{n-i}.$$
I think it is true that $$ b_k = \sum_{i=0}^k \binom{n-i}{n-k} a_{n-i}.$$ 
I am not sure how to prove this though. I tried induction but it gets quite complicated. It ends up having to confirm that for $m\leq n$ we have $$ \sum_{i=0}^{m-1} \binom{n-i}{n-m} a_{n-i} = \sum_{i=n-m+1}^m \sum_{j=0}^{n-i} (-1)^{i-n+m+1} \binom{n-j}{n-m, i-n+m, n-j-i} a_{n-j}$$ where the big bracket on the right is a trinomial coefficient. 
Can someone provide a proof please? Thanks.
 A: Using the formula for $a_k$, one can find the following relations:
$b_0 = a_N$
$b_1 = a_{N-1} - (-1)Nb_0$
$b_2 = a_{N-2} - (-1)Nb_1 - N(N-1)b_0$
and so on. Then, one can write all this in matricial form as
$\left(\begin{array}{c} b_0\\b_1\\\vdots\\b_N \end{array} \right) = \left( \begin{array}{c} a_N\\a_{N-1}\\\vdots\\a_0 \end{array}\right) + \Omega\left(\begin{array}{c} b_0\\b_1\\\vdots\\b_N \end{array}\right)$,
where $\Omega$ is defined as
$\Omega = \left( \begin{array}{ccccc} 0 & 0 & 0 &\dots & 0 \\ 
N & 0 & 0 & \dots & 0 \\
 -N(N-1) & N & 0 & \dots & 0 \\
\vdots & \vdots & \vdots & & \vdots
\end{array} \right)$
in its first rows (next ones are easily derived).
Given this, we can write the function for $b_i$ as
$\left(\begin{array}{c} b_0\\b_1\\\vdots\\b_N \end{array}\right) = (I-\Omega)^{-1}\left(\begin{array}{c} a_N\\a_{N-1}\\\vdots\\a_0 \end{array} \right) $,
where the only issue is to find the inverse of matrix $(I - \Omega)$. You can check the (in)validity of your hypothesis for $b_k$ (as it is posted in the comments, that formula is not valid).
