$\newcommand{\im}{\operatorname{im}}$
Given a complex $0\rightarrow V_0 \rightarrow V_1 \rightarrow \cdots \rightarrow V_r\rightarrow 0$, write $B_i=\im(V_{i-1} \rightarrow V_i)$ and $Z_i=\ker (V_i \rightarrow V_{i+1})$.
Show $\sum_{i=0}^{i=r} (-1)^i \dim V_i=\sum_{i=0}^{i=r} (-1)^i \dim H_i $, where $H_i=Z_i / B_i$ is the $i$th homology of the complex.
Here is my work:
Since $V_\cdot$ is a complex, $B_i \subset Z_i$.
Also, $\dim Z_i = \dim B_i + \dim H_i$.
Define $\phi_i:=V_{i-1} \rightarrow V_i$ and $\phi_{i+1}:=V_{i} \rightarrow V_{i+1}$. By rank-nullity theorem,
$$\dim Z_i +\dim(\im \phi_{i+1})=\dim V_i$$
$$\dim (\ker \phi_i) +\dim B_i=\dim V_i.$$
Combining three equations above, I need to show $\dim V_i=\dim (\ker \phi_i)-\dim(\im \phi_{i+1})$. If this approach works, could you give me any hint to continue?
(Thanks to @Michael Hardy for editing my question.)
I presume the $V_i$ are finite-dimensional vector spaces?
I do not see any exact sequences here.
$\newcommand{\im}{\operatorname{im}}$ You write both $$\dim Z_i +\dim(\im \phi_{i+1})=\dim V_i.$$ But this says $$\dim Z_i +\dim B_{i+1}=\dim V_i.\tag{$\dagger$}$$ You also say $$\dim (\ker \phi_i) +\dim B_i=\dim V_i.$$ But that is incorrect: the right side should be $\dim V_{i-1}$ which is $(\dagger)$ with $i$ replaced by $i-1$.
Now $(\dagger)$ implies $$\dim H_i+\dim B_i +\dim B_{i+1}=\dim V_i.\tag{$\ddagger$}$$ Multiply $(\ddagger)$ by $(-1)^i$ and add up over all $i$.