Homotopy Colimit of Truncations Let $\mathcal{A}$ be an additive category with countable coproducts. I am just starting to learn about homotopy colimits and I am struggling with the following example that I am very interested in understanding (you can find this as example 2 of section 4.4 in this paper): 
Given an $\mathcal{A}$-complex $X$, we let $X'$ denote the homotopy colimit of $$\tau_n X\to \tau_{n-1} X\to \tau_{n-2} X\to\ldots$$ where $$(\tau_n X)^p:=\left\{\begin{array}{cl} X^p & p\geq n\\ 0 & p<n\end{array}\right. ,$$  $\partial_{\tau_nM}^p=0$ for $p<n$ and $\partial_{\tau_nM}^p=\partial_{M}^p$ for $p\geq n$, and the maps are  just the inclusions of the truncations (I could be wrong about this, but I think these are the maps based on the paper).
The part I am having a hard time justifying is that $X'$ is isomorphic to $X$ in $K(\mathcal{A})$, the homotopy category of $\mathcal{A}$. Is there an easy way to see this, I honestly have no idea where to start when showing this? Thanks for any help!
 A: If $0\to A^\bullet\stackrel{\alpha}{\to}B^\bullet\stackrel{\beta}{\to}C^\bullet\to0$ is a sequence of complexes over $\mathcal{A}$ such that, in every degree $m$, the sequence
$0\to A^m\to B^m\to C^m\to0$ is a split short exact sequence, then there is a distinguished triangle $A^\bullet\stackrel{\alpha}{\to}B^\bullet\stackrel{\beta}{\to}C^\bullet\to A[1]$ in $K(\mathcal{A})$.
The homotopy colimit is defined by a triangle
$$\coprod_{i\leq n}\tau_iX\stackrel{1-\varphi}{\to}\coprod_{i\leq n}\tau_iX\to\operatorname{hocolim}\tau_iX\to\coprod_{i\leq n}\tau_iX[1]$$ 
where $\varphi$ is the map that is the inclusion $\tau_iX\to\tau_{i-1}X$ on the summand $\tau_iX$.
Consider the sequence
$$0\to\coprod_{i\leq n}\tau_iX\stackrel{1-\varphi}{\to}\coprod_{i\leq n}\tau_iX\stackrel{\theta}{\to}X\to0$$
where $\theta$ is the inclusion $\tau_iX\to X$ on the summand $\tau_iX$.
In degrees $m\geq n$ this gives
$$0\to\coprod_{i\leq n}X^m\stackrel{1-\varphi}{\to}\coprod_{i\leq n}X^m\stackrel{\theta}{\to}X^m\to0$$
and in degrees $m<n$ we get the same except that the coproduct is over $i\leq m$.
But this is a split short exact sequence, with splitting maps
$$X^m\to \coprod_{i\leq n}X^m$$
mapping by the identity to the first ($i=n$) summand, and
$$\coprod_{i\leq n}X^m\to\coprod_{i\leq n}X^m$$
which is zero on the first ($i=n$) summand, and on the $i=j<n$ summand is minus the sum of the identity maps to the summands for $i$ with $j<i\leq n$.
Maybe this is easier to follow when $\mathcal{A}$ is a module category, so we can describe the maps in terms of elements. Then (for $0=n\leq m$, say) this expresses $\coprod_{i\leq0}X^m$ as the direct sum of $\coprod_{i\leq0}X^m$ and $X^m$, with inclusion maps $$(x_0,x_{-1},x_{-2},\dots)\mapsto(x_0,x_{-1}-x_0,x_{-2}-x_{-1},\dots)$$
and
$$x\mapsto(x,0,0,\dots)$$
and projection maps
$$(y_0,y_{-1},y_{-2},\dots)\mapsto\left(-(y_{-1}+y_{-2}+y_{-3}+\dots),-(y_{-2}+y_{-3}+y_{-4}+\dots),\dots\right)$$
and
$$(y_0,y_{-1},y_{-2},\dots)\mapsto y_0+y_{-1}+y_{-2}+\dots.$$
