The 13 th term in the expansion of \(\left(x^{2}+\frac{2}{x}\right)^{n}\) is independent of \(x\), then the sum of the divisors of \(n\) is

13th term in the expansion of \(\left(x^{2}+\frac{2}{x}\right)^{n}\) is given by
\[
\begin{aligned}
\mathrm{T}_{13} &={ }^{\mathrm{n}} \mathrm{C}_{12}\left(\mathrm{x}^{2}\right)^{\mathrm{n}-12}\left(\frac{2}{\mathrm{x}}\right)^{12} \\
&={ }^{\mathrm{n}} \mathrm{C}_{12} \mathrm{x}^{2 \mathrm{n}-24}\left(\frac{2^{12}}{\mathrm{x}^{12}}\right) \\
&={ }^{\mathrm{n}} \mathrm{C}_{12} \mathrm{x}^{2 \mathrm{n}-24-12}\left(2^{12}\right) \\
&={ }^{\mathrm{n}} \mathrm{C}_{12} \mathrm{x}^{2 \mathrm{n}-36}\left(2^{12}\right)
\end{aligned}
\]
If the 13 th term will be independent of \(x\), then \(\mathrm{x}^{2 \mathrm{n}-36}\) must be \(0 .\)
i.e., \(2 \mathrm{n}-36=0 \Rightarrow 2 \mathrm{n}=36 \Rightarrow \mathrm{n}=18\)
The divisors of \(n=18\) are \(1,2,3,6,9,18\) and their sum \(=1+2+3+6+9+18=39\)