Proof that the Euler characteristic is additive I'm reading through a set of notes which assumes that the Euler characteristic is additive, but doesn't give a proof, so I would like to understand why this is.
Let $A_n$ be a finitely generated Abelian group, non-zero for $0<n<k$. Consider the long exact sequence
$ \cdots \rightarrow A_n \rightarrow B_n \rightarrow C_n \rightarrow A_{n-1} \rightarrow B_{n-1} \rightarrow C_{n-1} \rightarrow \cdots $
where $B_n$, $C_n$ defined similarly to $A_n$. 
I know that the Euler characteristic of $A$ is given by $\chi(A) = \sum_0^k (-1)^n \text{rank}(A_n)$, with similar definitions for $B$ and $C$. I want to prove that $\chi(B) = \chi(A) + \chi(C)$.
The notes say that for an exact sequence
$0 \rightarrow X \rightarrow Y \rightarrow Z \rightarrow 0$
of finitely generated Abelian groups $X,Y,Z$, we have the result rank($Y$) $=$ rank($X$) +  rank($Z$), and that the additivity of the Euler characteristic follows from this fact, but I can't see how.
 A: Hint. Decompose the exact sequence you start with with many short exact sequences, each one corresponding to the kernel and image of the maps in the first one.
For example, if $f:A\to B$ is a map, you can construct short exact sequences $$0\to\ker f\to A\to\operatorname{im}f\to0$$.
Later. The point of the hint was for you to use the observation provided by your notes themselves. $\def\rk{\operatorname{rk}}\def\im{\operatorname{im}}$Suppose that we have an exact complex $$0\xrightarrow{f_{-1}} X_0\xrightarrow{f_0} X_1\xrightarrow{f_1} X_2\xrightarrow{f_2}\cdots\xrightarrow{f_{n-1}} X_n\xrightarrow{f_n} 0$$ For each $i\in\{-1,\dots,n\}$ we have the map $f_i:X_i\to X_{i+1}$ (letting $X_{-1}=X_{n+1}=0$ for convenience) so we have a short exact sequence $$0\to\ker f_i\to X_i\to\im f_i\to 0$$ so, assuming we know that the rank is additive in short exact sequences, we have $$\rk X_i=\rk\ker f_i+\rk\im f_i.$$ Multiplying this by $(-1)^i$ and summing over $i$ we see then that $$\sum_{i=-1}^n(-1)^i\rk X_i=\sum_{i=-1}^n(-1)^i\rk\ker f_i+\sum_{i=-1}^n(-1)^i\rk\im f_i. \tag{1}$$ Now, the original exact sequence being exact, we have $\ker f_{i+1}\cong\im f_i$ for all $i$, so of course $\rk\ker f_{i+1}=\rk\im f_i$ for all $i$. Using this in the right hand side of (1) we easily see that in fact $$\sum_{i=-1}^n(-1)^i\rk X_i=0.$$
In your case, my $X_i$s are $A$s, $B$s and $C$s, so you just need to rename stuff
