= Solution
{c}
For each <spin> let $\mathcal F_\sigma$ contain the $N_\sigma$ lowest <tight binding> <energy levels>, and set $n_\sigma=N_\sigma/M$ and $C_\sigma=M^{-1}\sum_{k\in\mathcal F_\sigma}e^{ik}$. The normalized trial <Slater determinant> is
$$
|\mathcal F\rangle=\prod_{\sigma}\prod_{k\in\mathcal F_\sigma}\widetilde c^\dagger_{\sigma,k}|0\rangle.
$$
Its nearest-neighbour <one-body density matrix> is $\langle c^\dagger_{\sigma,j}c_{\sigma,j+1}\rangle=C_\sigma$. By <Wick's theorem>, <fermionic exchange symmetry> subtracts the squared off-diagonal <one-body density matrix> from the direct contribution
$$
\langle n_{\sigma,j}n_{\sigma,j+1}\rangle=n_\sigma^2-|C_\sigma|^2,
\qquad
\boxed{\langle H\rangle=M\sum_\sigma\left[-2t\operatorname{Re}C_\sigma+V(n_\sigma^2-|C_\sigma|^2)\right].}
$$
This formula also specifies the finite-$M$ answer exactly, including partially filled terminal <energy levels> through the chosen occupied set. For $N_\sigma$ consecutive allowed <wavevectors>, $|C_\sigma|=|\sin(\pi N_\sigma/M)|/[M\sin(\pi/M)]$; its phase locates the centre of that occupied interval.
Write $\tau=|t|$. In the large-$M$ limit the <Fermi sea> is centred at $k_0=0$ for $t>0$ and $k_0=\pi$ for $t<0$, with <Fermi wavevector> $k_{F,\sigma}=\pi n_\sigma$ measured from $k_0$. Consequently
$$
C_\sigma\longrightarrow e^{ik_0}\frac{\sin(\pi n_\sigma)}\pi,\qquad
\boxed{\frac{\langle H\rangle}{M}=\sum_{\sigma=\uparrow,\downarrow}\left[-\frac{2\tau}{\pi}\sin(\pi n_\sigma)+V\left(n_\sigma^2-\frac{\sin^2(\pi n_\sigma)}{\pi^2}\right)\right].}
$$
The subtraction due to <fermionic exchange symmetry> is essential: replacing $\langle n_jn_{j+1}\rangle$ by $n^2$ would discard the suppression of neighbouring equal-<spin> occupation. A completely filled <spin> band has zero <kinetic energy> and interaction <energy> $MV$, as the formula also shows.
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