1. Find the term that does not contain $x$ in the expansion of $\left(\frac{1}{x}+3 x\right)^{n}$, if the coefficient of the power of $x$ in the tenth term is the largest.
Solution. For the $(k+1)$-th term in the expansion, we get

$$
T_{k+1}=\binom{n}{k}\left(\frac{1}{x}\right)^{n-k}(3 x)^{k}=\binom{n}{k} 3^{k} x^{2 k-n}
$$

so the coefficient of $x$ is $A_{k+1}=\binom{n}{k} 3^{k}$. According to the condition of the problem, we have $A_{10}>A_{9}$ and $A_{10}>A_{11}$. From $A_{10}>A_{9}$, we get $\left({ }_{9}^{n}\right) 3^{9}>\left({ }_{8}^{n}\right) 3^{8}$, so $\frac{n!}{(n-9)!9!} 3^{9}>\frac{n!}{(n-8)!8!} 3^{8}$, which simplifies to $\frac{3}{9}>\frac{1}{n-8}$. From this, $3 n-24>9$, i.e., $n>11$. From $A_{10}>A_{11}$, we get $\binom{n}{9} 3^{9}>\binom{n}{10} 3^{10}$, and proceeding similarly as before, we get $n<12 \frac{1}{3}$.

Therefore, $n=12$.

Now $x^{2 k-12}=x^{0}$, so $2 k-12=0$, i.e., $k=6$. Thus, the desired term is the seventh, and it is $T_{6}=\left(\frac{12}{6}\right) 3^{6}=673596$.