The `13^(th)` term in the expanion of `(x^(2)+2//x)^(n)` is independent of `x` then the sum of the divisiors of `n` is

Options

A. `36`
B. `37`
C. `38`
D. `51`

Explanation

To solve the problem step by step, we need to find the 13th term in the expansion of \((x^2 + \frac{2}{x})^n\) and determine the value of \(n\) such that this term is independent of \(x\). Then, we will calculate the sum of the divisors of \(n\).

### Step 1: Identify the general term in the binomial expansion
The general term (the \(r+1\)th term) in the expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \(a = x^2\) and \(b = \frac{2}{x}\).

### Step 2: Write the expression for the 13th term
For the 13th term, we have \(r = 12\) (since \(r\) starts from 0):
\[ T_{13} = \binom{n}{12} (x^2)^{n-12} \left(\frac{2}{x}\right)^{12} \]
This simplifies to:
\[ T_{13} = \binom{n}{12} (x^2)^{n-12} \cdot \frac{2^{12}}{x^{12}} = \binom{n}{12} \cdot 2^{12} \cdot x^{2(n-12) - 12} \]

### Step 3: Simplify the exponent of \(x\)
Now, we simplify the exponent of \(x\):
\[ T_{13} = \binom{n}{12} \cdot 2^{12} \cdot x^{2n - 24 - 12} = \binom{n}{12} \cdot 2^{12} \cdot x^{2n - 36} \]

### Step 4: Set the exponent of \(x\) to zero
For the term to be independent of \(x\), the exponent of \(x\) must be zero:
\[ 2n - 36 = 0 \]
Solving for \(n\):
\[ 2n = 36 \implies n = 18 \]

### Step 5: Find the divisors of \(n\)
Now we need to find the sum of the divisors of \(n = 18\). The divisors of 18 are:
1, 2, 3, 6, 9, 18

### Step 6: Calculate the sum of the divisors
Now we sum these divisors:
\[ 1 + 2 + 3 + 6 + 9 + 18 = 39 \]

Thus, the sum of the divisors of \(n\) is \(39\).

### Final Answer
The sum of the divisors of \(n\) is \(39\).

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