= Solution
Writing $S=\sum X_i$, the <likelihood function> is $\theta^S(1-\theta)^{n-S}$ and its supremum occurs at $\widehat\theta=S/n$, including the limiting boundary cases zero and one. The <likelihood ratio> compares $2^{-n}$ under the null with that supremum. Thus
$$
\boxed{-2\log\Lambda=T_n=2n[\widehat\theta\log\widehat\theta+(1-\widehat\theta)\log(1-\widehat\theta)+\log2]}.
$$
Use $0\log0=0$ at boundary outcomes. <Wilks theorem> gives asymptotic $\chi^2_1$ under the regular interior null, because one parameter is constrained. Directly, setting $\widehat\theta=1/2+Z$ and expanding the bracket gives $2Z^2+O(Z^4)$, hence
$$
\boxed{T_n=4nZ_n^2+O(nZ_n^4)}.
$$
Under the null, $2\sqrt nZ_n\xrightarrow d N(0,1)$ and $Z_n=O_P(n^{-1/2})$. The remainder vanishes in probability, proving the same <chi-squared distribution> limit. If an observed $Z_n$ is not small, the quadratic numerical approximation is unreliable and the exact statistic should be used; exact binomial calibration is available. Persistent non-small deviations under a fixed alternative instead make the statistic grow linearly with $n$, rather than having the null <chi-squared distribution> law.
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