Explanation:
A Given, \(\left(x^2+\frac{2}{x}\right)^n\)
On comparing \((x+a)^n-\)
Here, \(\mathrm{x}=\mathrm{x}^2, \mathrm{a}=\frac{2}{\mathrm{x}}, \mathrm{n}=\mathrm{n}\)
Then, \(\mathrm{T}_{13}={ }^{\mathrm{n}} \mathrm{C}_{12}\left(\mathrm{x}^2\right)^{\mathrm{n}-12}\left(\frac{2}{\mathrm{x}}\right)^{12}\)
\(={ }^n C_{12} x^{(2 n-24)} \frac{(2)^{12}}{x^{12}}\)
\(={ }^n C_{12} x^{2 n-36} 2^{12}\)
Since, the \(13^{\text {th }}\) the terms is independent of \(x\).
Then, \(\mathrm{x}^{2 \mathrm{n}-36}=\mathrm{x}^0\)
\(2 \mathrm{n}-36=0\)
\(2 \mathrm{n}=36\)
\(\mathrm{n}=18\)
Then, the divisor of \(n=18\) are \(1,2,3,6,9,18\)
So, their sum \(=1+2+3+6+9+18=39\)