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Let $R$ be a Noetherian ring and $I$ a proper ideal of $R$. Suppose that the projective dimension of $I$ is equal to $n$. Let $0 \rightarrow P_n \rightarrow \cdots \rightarrow P_0 \rightarrow I \rightarrow 0$ be a projective resolution of $I$. Combining this resolution with the exact sequence $0 \rightarrow I \rightarrow R \rightarrow R /I \rightarrow 0$ we get a projective resolution $0 \rightarrow P_n \rightarrow \cdots \rightarrow P_0 \rightarrow R \rightarrow R/I \rightarrow 0$ of length $n+1$ for $R/I$. This implies that the projective dimension of $R/I$ can be at most $n+1$.

Question: Under what conditions do we have that $\operatorname{projdim} R/I = \operatorname{projdim} I + 1$ given that $I \neq R$?

Remark: This is a follow up question on my previous question isomorphic ideals and projective dimensions of quotients.

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If $0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0$ is an exact sequence of $R$-modules, and $\operatorname{pd}M'>\operatorname{pd}M$, then $\operatorname{pd}M''=\operatorname{pd}M'+1$. (This can be proved, for instance, by using the Ext characterization of projective dimension.)

For $0 \rightarrow I \rightarrow R \rightarrow R /I \rightarrow 0$, if $I$ is not projective, then $\operatorname{pd}R/I=\operatorname{pd}I+1$. Otherwise, there are two possibilities: $\operatorname{pd}R/I=0$ or $\operatorname{pd}R/I=1$ depending on whether $I$ is a direct summand of $R$ or not.