= Binomial inversion
{title2=$b_n=\sum_{k=0}^n(-1)^{n-k}\binom nk a_k\quad\Longleftrightarrow\quad a_n=\sum_{k=0}^n\binom nk b_k$}
These two finite-sum transformations are inverses over any <commutative ring>. Substitute one sum into the other and use $\binom nk\binom kj=\binom nj\binom{n-j}{k-j}$. The inner alternating sum is $(1-1)^{n-j}$ by the <binomial theorem>, so only $j=n$ survives. In a <Mahler expansion>, $b_n$ is the nth <finite difference> of the sampled values $a_k$.
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