= Solution
Write the end-to-end <random vector> as $\mathbf r=\sum_{j=1}^N b\mathbf u_j$, with independent isotropically distributed unit vectors. At zero applied <force>, $\langle\mathbf u_j\rangle=0$, $\mathbf u_j\cdot\mathbf u_j=1$, and for $i\ne j$ independence gives $\langle\mathbf u_i\cdot\mathbf u_j\rangle=0$. Therefore
$$
\langle\mathbf r\rangle=0,\qquad
\langle r^2\rangle=b^2\sum_{i,j}\langle\mathbf u_i\cdot\mathbf u_j\rangle=Nb^2,
\qquad\boxed{r_{\rm rms}=b\sqrt N}.
$$
Here “size” is the <root mean square> <end-to-end distance of a polymer chain>. It is not the magnitude of the mean <random vector>, which is zero, nor the mean distance $\langle r\rangle$. In the long-chain <Gaussian limit of the end-to-end distribution of a freely jointed chain>, $\langle r\rangle\simeq b\sqrt{8N/(3\pi)}$, illustrating why these definitions cannot be interchanged.
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