Let $f\colon B\rightarrow C$ be a map of filtered chain complexes. For each $r\geq 0$, define a filtration on the mapping cone $cone(f)$ by $F_p cone(f)=F_{p-r}B_{n-1}\oplus F_pC_n$. Show that $E_{p}^r(cone f)$ is the mapping cone of $f^r\colon E_p^r(B)\rightarrow E_p^r(C)$.
My attempt
I have managed to show that this is true when we are letting $r=0$. My thought was to try and use induction from here. However, I do not see how that can be helpful as the filtration on the cone changes when we change $r$. Because of this it does not seem like knowing the result is true for previous pages will give the result for the $r$th page. Just looking at the $1$st page, I am not even convinced the result is true since I think the objects for $E_p^1cone(f)$ should be $H_1\left(\frac{F_{p-1}B_{n-1}\oplus F_pC_n}{F_{p-2}B_{n-1}\oplus F_{p-1}C_n}\right)=H_1(E_{p-1,n-p}^0(B)\oplus E_{p,n-p}^0(C))=E_{p-1,n-p}^1(B)\oplus E_{p,n-p}^1(C)$ and the objects for the cone of $f^1$ should be $E_{p,n-1-p}^1(B)\oplus E_{p,n-p}^1(C)$.