= Solution
Put $\mathbf W_2=\mathbf F_2/\zeta$. To the needed accuracy the <Faxén translation law> gives $\mathbf U_2=\mathbf W_2+\mathsf M\mathbf V+O(\epsilon^3|\mathbf V|)$. Enforcing its prescribed <velocity> therefore gives
$$
\boxed{\mathbf F_2=6\pi\mu a(I-\mathsf M)\mathbf V
+O(6\pi\mu a\epsilon^3|\mathbf V|).}
$$
Here the <force> on sphere 1 remains prescribed; no extra second-order inversion term is needed in $\mathbf F_2$. Substitute this <force> into the incident field at sphere 1:
$$
\boxed{\mathbf U_1=(I+\mathsf M-\mathsf M^2)\mathbf V
+O(\epsilon^3|\mathbf V|),\qquad
\mathsf M^2=\frac{9a^2}{16R^2}(I+3\mathbf n\mathbf n).}
$$
The negative second-order term reflects the reduced <force> required on sphere 2 because sphere 1 already entrains it.
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