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.

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
