Solution
= Solution
Each leave-one-out statistic has the same distribution as $T_{n-1}$, so
$$
\mathbb E\widehat B_n
=(n-1)(B_{n-1}(\theta)-B_n(\theta)).
$$
Consequently the bias of the <jackknife bias correction> is
$$
\mathbb E\widetilde T_{\rm JACK}-\theta
=nB_n-(n-1)B_{n-1}.
$$
The assumed expansion gives
$$
nB_n=a+\frac bn+O(n^{-2})
$$
and
$$
(n-1)B_{n-1}
=a+\frac b{n-1}+O(n^{-2})
=a+\frac bn+O(n^{-2}).
$$
Their difference is therefore $O(n^{-2})$, as required.