The 13^{th} term in the expansion of left(x^{ | vedclass

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(D) The $13^{th}$ term in the expansion of $\left(x^{2}+\frac{2}{x}\right)^{n}$ is given by the general term formula $T_{r+1} = {}^{n}C_{r} a^{n-r} b^

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$39$

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(D) The $13^{th}$ term in the expansion of $\left(x^{2}+\frac{2}{x}\right)^{n}$ is given by the general term formula $T_{r+1} = {}^{n}C_{r} a^{n-r} b^{r}$.For $r = 12$,we have:$T_{13} = {}^{n}C_{12} (x^{2})^{n-12} (\frac{2}{x})^{12}$$T_{13} = {}^{n}C_{12} x^{2n-24} \cdot \frac{2^{12}}{x^{12}}$$T_{13} = {}^{n}C_{12} \cdot 2^{12} \cdot x^{2n-24-12}$$T_{13} = {}^{n}C_{12} \cdot 2^{12} \cdot x^{2n-36}$Since the term is independent of $x$,the exponent of $x$ must be $0$:$2n - 36 = 0$ $\Rightarrow 2n = 36$ $\Rightarrow n = 18$.The divisors of $n = 18$ are $1, 2, 3, 6, 9, 18$.The sum of the divisors is $1 + 2 + 3 + 6 + 9 + 18 = 39$.

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:

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