Let $$ \begin{array}{rccccl} A_0&\to& B_0&\to& C_0&\to\\ \downarrow & &\downarrow&&\downarrow\\ A_1&\to& B_1&\to& C_1&\to\\ \downarrow & &\downarrow&&\downarrow\\ \vdots & &\vdots&&\vdots\\ \end{array} $$ be a commutative diagram in a triangulated category such that all the rows are exact triangles. One defines the homotopy colimit of a sequence $A_0\to A_1\to\ldots$ as a third object in an exact triangle $$ \coprod A_n\xrightarrow{1-shift}\coprod A_n\to \operatorname{hocolim}_nA_* $$ which is unique up to (non-unique) isomorphism. For a morphism of diagrams $A_*\to B_*$ (as above), one gets a (non-unique) morphism $\operatorname{hocolim}_nA_*\to \operatorname{hocolim}_nB_*$.

Given the diagram above, is there a sequence of morphisms $$ \operatorname{hocolim}_nA_*\to \operatorname{hocolim}_nB_*\to \operatorname{hocolim}_nC_*\to $$ which is an exact triangle? I am especially interested in the case that $B_*$ is a constant diagram.

I was trying to use the 9-lemma for triangulated categories, but this explicitely **constructs** the morphisms $\coprod C_n\to \coprod C_n\to \tilde C$ and I cannot prove, why the first map has to be $1-shift$ and therefore why $\tilde C$ has to be the homotopy colimit of $C_0\to C_1\to\ldots$. This question is strongly related, I think.