The constant term in the expansion of `(x+(1)/(x^(3)))^(12)` is :

Options

A. 26
B. 169
C. 260
D. 220

Answer

To find the constant term in the expansion of \((x + \frac{1}{x^3})^{12}\), we can follow these steps:

### Step 1: Identify the General Term
The general term \(T_r\) in the expansion of \((a + b)^n\) is given by:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
In our case, \(n = 12\), \(a = x\), and \(b = \frac{1}{x^3}\). Thus, the general term becomes:
\[ T_r = \binom{12}{r} (x)^{12-r} \left(\frac{1}{x^3}\right)^r \]

### Step 2: Simplify the General Term
Now, we can simplify \(T_r\):
\[ T_r = \binom{12}{r} x^{12-r} \cdot \frac{1}{x^{3r}} = \binom{12}{r} x^{12 - r - 3r} = \binom{12}{r} x^{12 - 4r} \]

### Step 3: Find the Constant Term
The constant term occurs when the exponent of \(x\) is zero:
\[ 12 - 4r = 0 \]
Solving for \(r\):
\[ 4r = 12 \implies r = 3 \]

### Step 4: Substitute \(r\) to Find the Constant Term
Now, we substitute \(r = 3\) back into the general term:
\[ T_3 = \binom{12}{3} x^{12 - 4 \cdot 3} = \binom{12}{3} x^{0} = \binom{12}{3} \]

### Step 5: Calculate \(\binom{12}{3}\)
Now we calculate \(\binom{12}{3}\):
\[ \binom{12}{3} = \frac{12!}{3!(12-3)!} = \frac{12!}{3! \cdot 9!} \]
This simplifies to:
\[ \binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = \frac{1320}{6} = 220 \]

### Conclusion
Thus, the constant term in the expansion of \((x + \frac{1}{x^3})^{12}\) is \(220\).

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