If $x_1,...,x_n \in \mathfrak m$ is $M$-regular sequence, which $\operatorname{Tor}^R_i(M,R/(x_1,...,x_n))$ are $0$? Let $(R,\mathfrak m)$ be a Noetherian local ring. Let $M$ be a finitely generated $R$-module. If $x\in \mathfrak m$ is a non-zero-divisor on $M$, then it is easy to see $\operatorname{Tor}^R_1(M,R/(x))=0$.
My question is the following:

If $x_1,...,x_n \in \mathfrak m$ is an $M$-regular sequence, then is it true that $\operatorname{Tor}^R_i(M,R/(x_1,...,x_n))=0$ for all $i=1,...,n$ ? Or at least can we say $\operatorname{Tor}^R_1(M,R/(x_1,...,x_n))=0$, or $\operatorname{Tor}^R_n(M,R/(x_1,...,x_n))=0$ ?

I already know that the answer is affirmative if the sequence is also $R$-regular, but I do NOT want to assume that.
Here's the proof of $n=1$ case that I know: Consider the beginning of a free resolution of $R/(x)$: $F_2\to R \xrightarrow{\cdot x} R \to R/(x)\to 0$, where $F_2$ is some free module. To compute $\operatorname{Tor}^R_1(M,R/(x))$ we crop $R/(x)$ out of this and tensor with $M$ to get the complex $F_2\otimes_R M \to M \xrightarrow{\cdot x} M\to 0 $ , and the homology at the first spot is $\operatorname{Tor}^R_1(M,R/(x))$. But $x$ is $M$-regular, so $\operatorname{Tor}^R_1(M,R/(x))=\ker \{M \xrightarrow{\cdot x} M\}/Im \{F_2\otimes_R M \to M\}=0/Im \{F_2\otimes_R M \to M\}=0.$
Please help me with the general version.
Thank you.
 A: (1) It is true in general that $\operatorname{Tor}_1^R(M,R/(x_1,\dots,x_n))=0$.
To see this, consider the beginning of the Koszul complex for $(R,x_1,\dots,x_n)$:
$$
\bigwedge^2R^{\oplus n}\to R^{\oplus n}\xrightarrow{
\begin{pmatrix}
x_1&\dots&x_n
\end{pmatrix}
}R\to R/(x_1,\dots,x_n)\to 0.
$$
This is in general not exact at $R^{\oplus n}$, but if we add some free $R$-module $F$ to the leftmost term, we get an $R$-free resolution of $R/(x_1,\dots,x_n)$:
$$
F \oplus \bigwedge^2R^{\oplus n}\to R^{\oplus n}\to R\to R/(x_1,\dots,x_n)\to 0.
$$
Cutting out $R/(x_1,\dots,x_n)$ and tensoring with $M$, we obtain a complex
$$
(M\otimes_R F)\oplus \biggl(M\otimes_R\bigwedge^2R^{\oplus n}\biggr)\to M^{\oplus n}\to M
$$
whose cohomology at the middle gives $\operatorname{Tor}_1^R(M,R/(x_1,\dots,x_n)$.
Now, since $x_1,\dots,x_n$ is an $M$-regular sequence, the Koszul complex for $(M,x_1,\dots,x_n)$ is exact.
Therefore the above complex is exact at the middle, so we get $\operatorname{Tor}_1^R(M,R/(x_1,\dots,x_n))=0$.
(2) There is an example such that $\operatorname{Tor}_2^R(M,R/(x_1,x_2))\neq 0$.
Namely, let $k$ be a field and set $R=k[X,Y,Z]_{(X,Y,Z)}/(XY,XZ)$, $M=R/(X)\simeq k[Y,Z]_{(Y,Z)}$.
Then $Y,Z$ is obviously an $M$-regular sequence, but it is not an $R$-regular sequence.
One can check that the following sequence gives an $R$-free resolution of $M$:
$$
R^{\oplus 3}
\xrightarrow{
\begin{pmatrix}
Z&X&0\\
-Y&0&X
\end{pmatrix}
}
R^{\oplus 2}
\xrightarrow{
\begin{pmatrix}
Y&Z
\end{pmatrix}
}
R\xrightarrow{X}R\to M\to 0.
$$
Cutting out $M$ and tensoring with $R/(Y,Z)$, we obtain a complex
$$
R/(Y,Z)^{\oplus 3}\xrightarrow{
\begin{pmatrix}
0&X&0\\
0&0&X
\end{pmatrix}
}
R/(Y,Z)^{\oplus 2}\xrightarrow{
\begin{pmatrix}
0&0
\end{pmatrix}
}
R/(Y,Z)\xrightarrow{X}R/(Y,Z)
$$
whose cohomology at $R/(Y,Z)^{\oplus 2}$ gives $\operatorname{Tor}_2^R(M,R/(Y,Z))$.
Therefore we get $\operatorname{Tor}_2^R(M,R/(Y,Z))\simeq k^{\oplus 2}\neq 0$.
