When approximately dealing with organic polymer molecules, one approach is to focus attention on the skeleton structure composed of its carbon atoms. Assume that on a one-dimensional carbon backbone chain, there are only two possible connections between two carbon atoms: one is a long bond with a bond length of $a$ and energy $\varepsilon_{a}$; the other is a short bond with a bond length of $b$ and energy $\varepsilon_{b}$. If this one-dimensional chain contains $N$ carbon atoms, with an end-to-end distance of $L$ and tension $\mathbf{X}$, attempt to find what is its partition function and the average length of the molecule $\overline{{{l}}}$ ?

Options:
A. $$
[ e^{\beta( X_{a}-\varepsilon_{a} )}+e^{\beta( X_{b}-\varepsilon_{b} )} ]^{N}
$$
$$
\overline{{{{l}}}}=\frac{\overline{{{L}}}} {N} \sim\frac{a+b \mathrm{e}^{\delta}} {1+\mathrm{e}^{\delta}}+\frac{X} {k T} \biggl[ \frac{a^{2}+b^{2} \mathrm{e}^{\delta}} {1+\mathrm{e}^{\delta}}-\Bigl( \frac{a+b \mathrm{e}^{\delta}} {1+\mathrm{e}^{\delta}} \Bigr)^{2} \biggr]
$$
B. $$
[ e^{\beta( \epsilon_{b}-X_{a} )}+e^{\beta( \epsilon_{a}-X_{b} )} ]^{N}
$$
$$
\overline{{{{l}}}}=\frac{\overline{{{L}}}} {N} \sim\frac{a \mathrm{e}^{\delta}+b} {1+\mathrm{e}^{\delta}}+\frac{X} {k T} \biggl[ \frac{a^{2} \mathrm{e}^{\delta}+b^{2}} {1+\mathrm{e}^{\delta}}-\Bigl( \frac{a \mathrm{e}^{\delta}+b} {1+\mathrm{e}^{\delta}} \Bigr)^{2} \biggr]
$$
C. $$
[ e^{\beta( X_{a}-X_{b} )}+e^{\beta( \epsilon_{b}-\epsilon_{a} )} ]^{N}
$$
$$
\overline{{{{l}}}}=\frac{\overline{{{L}}}} {N} \sim\frac{a+b} {1+\mathrm{e}^{\delta}}+\frac{X} {k T} \biggl[ \frac{a^{2}+b^{2}} {1+\mathrm{e}^{\delta}}-\Bigl( \frac{a+b} {1+\mathrm{e}^{\delta}} \Bigr)^{2} \biggr]
$$
D. $$
[ e^{\beta( X_{b}-\epsilon_{a} )}+e^{\beta( X_{a}-\epsilon_{b} )} ]^{N}
$$
$$
\overline{{{{l}}}}=\frac{\overline{{{L}}}} {N} \sim\frac{b \mathrm{e}^{\delta}+a} {1+\mathrm{e}^{\delta}}+\frac{X} {k T} \biggl[ \frac{b^{2} \mathrm{e}^{\delta}+a^{2}} {1+\mathrm{e}^{\delta}}-\Bigl( \frac{b \mathrm{e}^{\delta}+a} {1+\mathrm{e}^{\delta}} \Bigr)^{2} \biggr]
$$
E. $$
[ e^{\beta( X_{b}-\epsilon_{b} )}+e^{\beta( X_{a}-\epsilon_{a} )} ]^{N}
$$
$$
\overline{{{{l}}}}=\frac{\overline{{{L}}}} {N} \sim\frac{b+a \mathrm{e}^{\delta}} {1+\mathrm{e}^{\delta}}+\frac{X} {k T} \biggl[ \frac{b^{2}+a^{2} \mathrm{e}^{\delta}} {1+\mathrm{e}^{\delta}}-\Bigl( \frac{b+a \mathrm{e}^{\delta}} {1+\mathrm{e}^{\delta}} \Bigr)^{2} \biggr]
$$
F. $$
[ e^{\beta( \epsilon_{b}-\epsilon_{a} )}+e^{\beta( X_{a}-X_{b} )} ]^{N}
$$
$$
\overline{{{{l}}}}=\frac{\overline{{{L}}}} {N} \sim\frac{a+b \mathrm{e}^{\delta}} {1-\mathrm{e}^{\delta}}+\frac{X} {k T} \biggl[ \frac{a^{2}+b^{2} \mathrm{e}^{\delta}} {1-\mathrm{e}^{\delta}}-\Bigl( \frac{a+b \mathrm{e}^{\delta}} {1-\mathrm{e}^{\delta}} \Bigr)^{2} \biggr]
$$
G. $$
[ e^{\beta( X_{b}-X_{a} )}+e^{\beta( \epsilon_{b}-\epsilon_{a} )} ]^{N}
$$
$$
\overline{{{{l}}}}=\frac{\overline{{{L}}}} {N} \sim\frac{a+b \mathrm{e}^{\delta}} {2}+\frac{X} {k T} \biggl[ \frac{a^{2}+b^{2} \mathrm{e}^{\delta}} {2}-\Bigl( \frac{a+b \mathrm{e}^{\delta}} {2} \Bigr)^{2} \biggr]
$$
H. $$
[ e^{\beta( X_{b}-\epsilon_{a} )}+e^{\beta( X_{a}-\epsilon_{b} )} ]^{N}
$$
$$
\overline{{{{l}}}}=\frac{\overline{{{L}}}} {N} \sim\frac{a+b \mathrm{e}^{-\delta}} {1+\mathrm{e}^{-\delta}}+\frac{X} {k T} \biggl[ \frac{a^{2}+b^{2} \mathrm{e}^{-\delta}} {1+\mathrm{e}^{-\delta}}-\Bigl( \frac{a+b \mathrm{e}^{-\delta}} {1+\mathrm{e}^{-\delta}} \Bigr)^{2} \biggr]
$$
I. $$
[ e^{\beta( \epsilon_{a}-X_{b} )}+e^{\beta( \epsilon_{b}-X_{a} )} ]^{N}
$$
$$
\overline{{{{l}}}}=\frac{\overline{{{L}}}} {N} \sim\frac{b+a \mathrm{e}^{\delta}} {1+\mathrm{e}^{\delta}}+\frac{k T} {X} \biggl[ \frac{b^{2}+a^{2} \mathrm{e}^{\delta}} {1+\mathrm{e}^{\delta}}-\Bigl( \frac{b+a \mathrm{e}^{\delta}} {1+\mathrm{e}^{\delta}} \Bigr)^{2} \biggr]
$$
J. $$
[ e^{\beta( \epsilon_{b}-X_{b} )}+e^{\beta( \epsilon_{a}-X_{a} )} ]^{N}
$$
$$
\overline{{{{l}}}}=\frac{\overline{{{L}}}} {N} \sim\frac{a+b \mathrm{e}^{2\delta}} {1+\mathrm{e}^{2\delta}}+\frac{X} {k T} \biggl[ \frac{a^{2}+b^{2} \mathrm{e}^{2\delta}} {1+\mathrm{e}^{2\delta}}-\Bigl( \frac{a+b \mathrm{e}^{2\delta}} {1+\mathrm{e}^{2\delta}} \Bigr)^{2} \biggr]
$$
Answer: A