How can I prove this relation?
${ a }_{ n }=\sum _{ k=0 }^{ n }{ { (-1) }^{ k }\left( \begin{matrix} n \\ k \end{matrix} \right) { b }_{ k } } \quad$ if and only if $\quad{ b }_{ n }=\sum _{ k=0 }^{ n }{ { (-1) }^{ k }\left( \begin{matrix} n \\ k \end{matrix} \right) a_{ k } } $
I think I have to consider a combinatoric proof... but I can't go any further... So, I searched about this topic, and can find some documents related to "binomial inverse pairs". But still I can't solve it.
To prove the corrected formula, put $\displaystyle f(x)=\sum_{n\geq 0} \frac{a_n}{n!}x^n$ and $\displaystyle g(x)=\sum_{n\geq 0} \frac{b_n}{n!}x^n$; then the first relation say that $f(-x)=\exp(-x)g(x)$, it is easy to finish.