= Solution
Let the applied <force> be $\mathbf f=f\hat{\mathbf z}$. A segment with polar angle $\theta$ has energy $-fb\cos\theta$. Independent 3D orientations give the one-link <partition function>
$$
Z_1=\int e^{fb\cos\theta/(k_BT)}d\Omega
=4\pi\frac{\sinh u}{u},\qquad u=\frac{fb}{k_BT},\qquad Z_N=Z_1^N.
$$
Taking the <force> <derivative> of $k_BT\ln Z_N$ yields the mean end-to-end projection,
$$
\langle R_z\rangle=Nb\left(\coth u-\frac1u\right).
$$
For $u\ll1$, $\coth u-1/u=u/3+O(u^3)$, so
$$
\boxed{\langle\mathbf R\rangle=\frac{Nb^2}{3k_BT}\mathbf f+O(f^3),\qquad
\mathbf f=\frac{3k_BT}{Nb^2}\langle\mathbf R\rangle.}
$$
<Isotropy> supplies the <vector> relation. It is the mean <vector> or its projection, not the mean magnitude of a fluctuating chain; at zero <force> $\langle\mathbf R\rangle=0$ while $\langle R^2\rangle=Nb^2$ remains positive. The last panel of the model figure in <past-exam-of-the-biological-physics-course-of-the-university-of-oxford/2019/4/b/solution> shows the exact orientational response and its weak-<force> tangent.
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