= Solution
The <coherent state> in the zero mode has mean <occupation number> $N_0$. The other <occupation numbers> follow from part (c), giving exactly
$$
\boxed{\langle\hat N\rangle=N_0+D,\qquad
D\equiv\sum_{\mathbf k\ne0}\sinh^2\theta_k.}
$$
Impose the fixed mean <particle number> $\langle\hat N\rangle=N$ by choosing
$$
\boxed{N_0=N-D.}
$$
Here $D$ is the <quantum depletion of a Bose condensate>. The approximation requires $0\le D\ll N$, so $N_0$ remains positive and macroscopically large. The specified <coherent state> and <multimode squeezed vacuum> are not exact <eigenstates> of the total <number operator>; $N$ in this variational construction is their prescribed mean <particle number>.
Write $P=\sum_{\mathbf k\ne0}s_kc_k$. Substituting into the retained part of the <interaction energy> before truncating gives
$$
\begin{aligned}
\langle H_{\rm int}\rangle
&\simeq\frac{u}{2V}(N-D)^2+\frac{u}{V}(N-D)(2D-P)\\
&=\frac{uN^2}{2V}+\frac{uN}{V}(D-P)
+\frac{u}{V}\left(-\frac32D^2+DP\right).
\end{aligned}
$$
The last two terms quadratic in $D,P$ have no explicit large-$N$ factor when the <Bogoliubov transformation> parameters are held fixed, and are discarded along with the nonzero-mode quartic contribution already omitted in part (c). This is the same truncation that underlies the <number-conserving Bogoliubov quadratic Hamiltonian>. Thus, defining the physical <number density> correctly as $\rho=N/V=N/L^3$,
$$
\boxed{\langle H_{\rm int}\rangle\simeq\frac{u\rho}{2}N
+u\rho\sum_{\mathbf k\ne0}
(\sinh^2\theta_k-\sinh\theta_k\cosh\theta_k).}
$$
Replacing $N_0$ by $N$ at the outset would give the wrong coefficient $2\sinh^2\theta_k$: expanding the leading condensate <interaction energy> supplies the necessary subtraction.
\b[The printed definition $\rho=N/L$ is an error.] In three dimensions the <contact interaction> coupling $u$ has units of <energy> times <volume>, so $uN/L$ does not even have units of <energy>. There is a direct counterexample with no depletion: when $\theta_k=0$, the specified <coherent state> has exactly $\langle H_{\rm int}\rangle=uN^2/(2L^3)$. In fixed length and <energy> units, $L=2$, $u=1$ and $N=100$ give $625$, whereas the printed $u\rho N/2$ with $\rho=N/L$ gives $2500$. The requested formula and the minimization in part (e) hold with $\rho=N/L^3$.
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