Two sequence relations are equivalent. For two sequences of complex numbers $\{a_0, a_1, ..., a_n, ...\}$ and $\{b_0, b_1, ..., b_n, ...\}.$ Show that the following relations are equivalent:
\begin{equation}
a_n = \sum_{k=0}^n b_k \iff b_n = \sum_{k=0}^n(-1)^{k+n}a_k \text{ for all } n.
\end{equation} 
I know it had to do with some type of Mobius inversion. I tried as follows
\begin{align}
\sum_{k=0}^n b_k &= \sum_{k=0}^n \sum_{m=0}^k(-1)^{m+k}a_m \\
                 &= \sum_{m=0}^n \sum_{k=m}^n(-1)^{m+k}a_m \\
                 &= \sum_{m=0}^n(-1)^ma_m \sum_{k=m}^n(-1)^{k} 
\end{align} 
I have no idea what to do next. I am not even sure that I interchange sum in correct way. Any help will be really appreciated. Thanks.
Note--I saw similar questions but they are not of so much help.
 A: I agree the question is not correct as stated. However, I believe it's a typo on the starting value of the first sum, with it meant to be for $k = n - 1$, as this will then make the question correct. Thus, the statement to prove would instead be
$$a_n = \sum_{k=n-1}^n b_k \iff b_n = \sum_{k=0}^n(-1)^{k+n}a_k \text{ for all } n \tag{1}\label{eq1}$$
This can be shown using induction. First, note that $n = 0$, where the sum starts at $0$, we get $a_0 = b_0$ on the left & right sides. For $n = 1$, the left side gives $a_1 = b_0 + b_1$, from which $b_1 = -b_0 + a_1 = -a_0 + a_1$, which is the right side.
Next, assume this holds true for all $n \le m$ for some integer $m \ge 0$. For $n = m + 1$, the left side of \eqref{eq1} becomes
$$a_{m+1} = b_{m} + b_{m+1} \Rightarrow b_{m+1} = -b_{m} + a_{m+1} \tag{2}\label{eq2}$$
From the induction hypothesis, we get that
\begin{align}
-b_{m} & = -\sum_{k=0}^m (-1)^{k+m}a_k \\
& = \sum_{k=0}^m (-1)^{k+\left(m + 1\right)}a_k \\
& = \sum_{k=0}^{m+1} (-1)^{k+\left(m + 1\right)}a_k - \left(-1\right)^{\left(m+1\right)+\left(m+1\right)}a_{m+1} \\
& = \sum_{k=0}^{m+1} (-1)^{k+\left(m + 1\right)}a_k - a_{m+1}
\end{align}
Substituting this into \eqref{eq2} gives that
$$b_{m+1} = \sum_{k=0}^{m+1} (-1)^{k+\left(m + 1\right)}a_k  \tag{3}\label{eq3}$$
This proves the statement one way. As all of the steps are reversible, this means the statement is also true the other way. Thus, it's also true for $n = m + 1$, completing the induction procedure.
