{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 $a_k$.

Binomial inversion

Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 27 / 2 / i / Solution

Binomial inversion ( $b_{n} = \sum_{k = 0}^{n} (- 1)^{n - k} (k n) a_{k} ⟺ a_{n} = \sum_{k = 0}^{n} (k n) b_{k}$ )

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Binomial inversion (bn​=∑k=0n​(−1)n−k(kn​)ak​⟺an​=∑k=0n​(kn​)bk​)

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Binomial inversion ( $b_{n} = \sum_{k = 0}^{n} (- 1)^{n - k} (k n) a_{k} ⟺ a_{n} = \sum_{k = 0}^{n} (k n) b_{k}$ )

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