Consider the expansion `( x^(2) + ( 1)/( x))^(15)`
What is the independent term in the given expansion ?

Options

A. 2103
B. 3003
C. 4503
D. None of these

Answer

To find the independent term in the expansion of \((x^2 + \frac{1}{x})^{15}\), we will follow these steps:

### Step 1: Identify the general term in the binomial expansion
The general term \(T_{r+1}\) in the binomial expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \(a = x^2\), \(b = \frac{1}{x}\), and \(n = 15\). Therefore, the general term becomes:
\[ T_{r+1} = \binom{15}{r} (x^2)^{15-r} \left(\frac{1}{x}\right)^r \]

### Step 2: Simplify the general term
Now, simplify the general term:
\[ T_{r+1} = \binom{15}{r} (x^{2(15-r)}) \left(x^{-r}\right) \]
This simplifies to:
\[ T_{r+1} = \binom{15}{r} x^{30 - 2r - r} = \binom{15}{r} x^{30 - 3r} \]

### Step 3: Find the condition for the independent term
To find the independent term, we need the exponent of \(x\) to be zero:
\[ 30 - 3r = 0 \]
Solving for \(r\):
\[ 30 = 3r \quad \Rightarrow \quad r = 10 \]

### Step 4: Substitute \(r\) back into the general term
Now, substitute \(r = 10\) into the general term:
\[ T_{11} = \binom{15}{10} x^{30 - 3(10)} = \binom{15}{10} x^0 \]
This means that the term is independent of \(x\).

### Step 5: Calculate \(\binom{15}{10}\)
Using the formula for combinations:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
We calculate:
\[ \binom{15}{10} = \frac{15!}{10! \cdot 5!} \]
This can be simplified as:
\[ \binom{15}{10} = \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} \]

### Step 6: Perform the calculations
Calculating the numerator:
\[ 15 \times 14 = 210 \]
\[ 210 \times 13 = 2730 \]
\[ 2730 \times 12 = 32760 \]
\[ 32760 \times 11 = 360360 \]
Now, calculating the denominator:
\[ 5! = 120 \]
Thus:
\[ \binom{15}{10} = \frac{360360}{120} = 3003 \]

### Conclusion
The independent term in the expansion of \((x^2 + \frac{1}{x})^{15}\) is:
\[ \boxed{3003} \]