Question

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:

• A. $36$
• B. $37$
• C. $38$
• D. $39$

Answer 1

The correct option is D

$39$

Explanation for the correct option:

Calculating the sum of the divisors of $n$:

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

$$\begin{gathered} =1+2+3+6+9+18 \\ =39 \end{gathered}$$

Hence, option (D) is the correct answer.