Obtaining derived functors from derived functors of similar complexes or "bluntly truncated" unbounded complexes (without adding 0's to the left) I don't know if I'm actually using the right terminology here, to be clear I'm going to state explicitly what I'm trying to figure out to see if I can be pointed in the right direction:
Let $F: \mathcal{A} \rightarrow \mathcal{B}$ be a left exact functor between 2 abelian categories $\mathcal{A}$ and $\mathcal{B}$, let $X^\bullet$ be the complex $0 \rightarrow X^0 \rightarrow X^1 \rightarrow X^2 \rightarrow X^3 \rightarrow \cdots$ of objects in $\mathcal{A}$, let $T^\bullet$ be a complex obtained by removing the FIRST TERM (or maybe the first $q$ terms from the left?), I mean let $T^\bullet$ be $X^0 \rightarrow X^1 \rightarrow X^2 \rightarrow X^3 \rightarrow \cdots$, the complex obtained by removing the $0$ from $X^\bullet$, NO 0's to the left, so $T$ becomes "unbounded". Let's say I have a way of obtaining the derived functor $RF(T^\bullet)$ using the machinery developed for unbounded complexes. Is there any way I can compute some of the terms in $RF(T^\bullet)$ from knowing this?
 A: Suppose for simplicity $T$ and $S$ are two bounded complexes supported in negative degrees which differ only in their $0$th term. Construct Cartan-Eilenberg projective resolutions $P_T$ and $P_S$ for them, so that $RF(T)$ and $RF(S)$ are in fact $F(\operatorname{Tot} P_T)$ and $F(\operatorname{Tot} P_S)$. Now, in view of the way one constructs C-E resolutions, $P_T$ and $P_S$ can be chosen in such a way that they only differ in specific places (the terms appearing in the $0$th column and in the differentials which have those terms as codomains, I think). If the global dimension $d$ of $\mathcal A$ is finite, and if one uses finite resolutions for objects of length at most $d$,  the complexes $F(\operatorname{Tot} P_T)$ and $F(\operatorname{Tot} P_S)$ therefore differ in at most $d+1$ places.
A: $T$ is called the stupid truncation of $X$ (in a contrast with the canonical or smart truncation).
In the derived category $D(A)$ there is a distinguished triangle
$$
T \to X \to R,
$$
where $R$ is the complex consisted of the ONE term (or $q$ terms) of $X$. Applying the derived functor $RF$ you get a distinguished triangle
$$
RF(T) \to RF(X) \to RF(R)
$$
in $D(B)$. So, $RF(R)$ measures the difference between $RF(T)$ and $RF(X)$. In particular there is a long exact sequence
$$
\dots \to R^nF(T) \to R^nF(X) \to R^nF(R) \to R^{n+1}F(T) \to \dots
$$
and if you know $R^nF(R)$ you can compare $R^nF(X)$ and $R^nF(T)$.
