12. When $n \leqslant 4$, it can be directly verified that the proposition holds. When $n \geqslant 5$, let $p_{1}, p_{2}$, $p_{3}, \cdots$ represent the sequence of all prime numbers in ascending order, and set $p_{n}=2 m+1$. It is known that
$$p_{1}+p_{2}+\cdots+p_{n}<1+3+5+7+\cdots+(2 m+1)=(m+1)^{2},$$
i.e., $a_{n}<(m+1)^{2}$. Now, assume $k^{2} \leqslant a_{n}<(k+1)^{2}$, then $k \leqslant m$. Also, $p_{n+1} \geqslant 2 m+3$, thus,
$$(k+1)^{2}=\dot{k}^{2}+2 k+1 \leqslant k^{2}+2 m+1<k^{2}+p_{n+1} \leqslant a_{n}+p_{n+1}=a_{n+1}$$

Therefore, $(k+1)^{2} \in\left[a_{n}, a_{n+1}\right]$.