If $a_n = \sum\limits_{r=0}^n \binom{n}{r} b_r$, prove $(-1)^n b_n = \sum\limits_{s=0}^n \binom{n}{s} (-1)^s a_s$ Suppose that sequences of real numbers satisfy:
\begin{align*}
  a_n &= \sum\limits_{r=0}^n \binom{n}{r} b_r \\
\end{align*}
Prove that:
\begin{align*}
  (-1)^n b_n &= \sum\limits_{s=0}^n \binom{n}{s} (-1)^s a_s \\
\end{align*}
My work:
The $n=0$ case:
\begin{align*}
  a_0 &= b_0 \\
  b_0 &= a_0 \\
\end{align*}
Then $n=1$ case:
\begin{align*}
  a_1 &= \binom{1}{0} b_0 + \binom{1}{1} b_1 = b_0 + b_1 \\
  -b_1 &= \binom{1}{0} a_0 - \binom{1}{1} a_1 = a_0 - a_1 \\
  b_1 &= a_1 - a_0 = b_0 + b_1 - a_0 = b_1 \\
\end{align*}
The inductive step:
\begin{align*}
  (-1)^{n+1} b_{n+1} &= \sum\limits_{s=0}^{n+1} \binom{n+1}{s} (-1)^s a_s \\
\end{align*}
I'm not sure where to go from here. I tried substituting $a_s$ into that last equation and it didn't help. Any ideas?
 A: See binomial transform for more details. Here's a proof using (exponential) generating series.
Let
$$ B(x) = \sum_{n = 0}^\infty b_n \frac{x^n}{n!} $$
be the exponential generating series for $(b_n)$.
Then
\begin{align}
e^x B(x) &= \left( \sum_{i = 0}^\infty \frac{x^i}{i!} \right)\left( \sum_{j = 0}^\infty b_j \frac{x^j}{j!} \right) \\
&= \sum_{n = 0}^\infty \left( \sum_{i + j = n} b_j \frac{x^i}{i!}\frac{x^j}{j!} \right) \\
&= \sum_{n = 0}^\infty \left( \sum_{k = 0}^n b_k \frac{x^{n - k}}{(n - k)!}\frac{x^{k}}{k!} \right) \\
&= \sum_{n = 0}^\infty \left( \sum_{k = 0}^n b_k \frac{n!}{(n - k)!k!} \right) \frac{x^{n}}{n!} \\
&= \sum_{n = 0}^\infty a_n \frac{x^n}{n!} = A(x).
\end{align}
Where $A(x)$ is the exponential generating series for $(a_n)$. That is,
$$ \color{purple}{A(x) = e^x B(x)} $$
and conversely,
$$ B(x) = e^{-x} A(x). \tag{$*$} $$
The product $e^{-x} B(x)$ can be written as
\begin{align}
e^{-x} A(x) &= \left( \sum_{i = 0}^\infty (-1)^{i}\frac{x^i}{i!} \right)\left( \sum_{j = 0}^\infty a_j \frac{x^j}{j!} \right) \\
&= \sum_{n = 0}^\infty \left( \sum_{i + j = n} (-1)^i a_j \frac{x^i}{i!}\frac{x^j}{j!} \right) \\
&= \sum_{n = 0}^\infty \left( \sum_{k = 0}^n (-1)^{n - k}a_k \frac{x^{n - k}}{(n - k)!}\frac{x^{k}}{k!} \right) \\
&= \sum_{n = 0}^\infty \left( (-1)^n \sum_{k = 0}^n (-1)^{k}a_k \frac{n!}{(n - k)!k!} \right) \frac{x^{n}}{n!} \\
&= B(x) = \sum_{n = 0}^\infty b_n \frac{x^n}{n!} \tag{by ($*$)}
\end{align}
Comparing the coefficient of $x^n/n!$ on both sides in $(*)$ gives us
$$ b_n = (-1)^n \sum_{k = 0}^n (-1)^{k}a_k \frac{n!}{(n - k)!k!} $$
or equivalently,
$$ \color{blue}{(-1)^n b_n = \sum_{k = 0}^n \binom{n}{k}(-1)^{k}a_k.} $$
A: Here is a direct proof
that uses only
$\sum\limits_{s=0}^{m} (-1)^{s}\binom{m}{s}
=1$
when $m=0$
and
$=0$ when
$m \ge 1$,
This last is
from the expansion
$0
=(1-1)^m
=\sum\limits_{s=0}^{m} (-1)^{s}\binom{m}{s}
$.
$\begin{array}\\
\sum\limits_{s=0}^n \binom{n}{s} (-1)^s a_s 
&=\sum\limits_{s=0}^n \binom{n}{s} (-1)^s 
\sum\limits_{r=0}^s \binom{s}{r} b_r \\
&=\sum\limits_{r=0}^n\sum\limits_{s=r}^n \binom{n}{s} (-1)^s 
 \binom{s}{r} b_r \\
&=\sum\limits_{r=0}^nb_r\sum\limits_{s=r}^n \binom{n}{s} (-1)^s 
 \binom{s}{r} \\
&=\sum\limits_{r=0}^nb_r\sum\limits_{s=r}^n (-1)^s \binom{n}{s} 
 \binom{s}{r} \\
&=\sum\limits_{r=0}^nb_r\sum\limits_{s=r}^n (-1)^s \dfrac{n!s!}{s!(n-s)!r!(s-r)!}\\
&=\sum\limits_{r=0}^nb_r\sum\limits_{s=r}^n (-1)^s \dfrac{n!}{(n-s)!r!(s-r)!}\\
&=\sum\limits_{r=0}^nb_r\dfrac{n!}{r!}\sum\limits_{s=r}^n (-1)^s \dfrac{1}{(n-s)!(s-r)!}\\
&=\sum\limits_{r=0}^nb_r\dfrac{n!}{r!(n-r)!}\sum\limits_{s=r}^n (-1)^s \dfrac{(n-r)!}{(n-s)!(s-r)!}\\
&=\sum\limits_{r=0}^nb_r\dfrac{n!}{r!(n-r)!}\sum\limits_{s=0}^{n-r} (-1)^{s+r} \dfrac{(n-r)!}{(n-(s+r))!(s+r-r)!}\\
&=\sum\limits_{r=0}^nb_r\binom{n}{r}\sum\limits_{s=0}^{n-r} (-1)^{s+r} \dfrac{(n-r)!}{(n-s-r)!s!}\\
&=\sum\limits_{r=0}^nb_r(-1)^r\binom{n}{r}\sum\limits_{s=0}^{n-r} (-1)^{s}\binom{n-r}{s}\\
&=(-1)^nb_n\\
\end{array}
$
To rephrase the last statement
in terms of the sum,
$\sum\limits_{s=0}^{n-r} (-1)^{s}\binom{n-r}{s}
=1$
when $n=r$
and
$=0$ when
$n > r$.
