The constant term in the expansion of `(1+ x+ (2)/(x))^(6)` is

Options

A. 479
B. 517
C. 569
D. 581

Answer

To find the constant term in the expansion of \((1 + x + \frac{2}{x})^6\), we can use the multinomial expansion. The general term in the expansion can be expressed as:

\[ T = \frac{6!}{a!b!c!} (1)^a (x)^b \left(\frac{2}{x}\right)^c \]

where \(a + b + c = 6\), and \(a\), \(b\), and \(c\) are the powers of \(1\), \(x\), and \(\frac{2}{x}\) respectively.

### Step 1: Identify the general term
The general term can be rewritten as:

\[ T = \frac{6!}{a!b!c!} (1)^a (x)^b \left(\frac{2}{x}\right)^c = \frac{6!}{a!b!c!} \cdot 1^a \cdot 2^c \cdot x^{b-c} \]

### Step 2: Find the condition for the constant term
For \(T\) to be a constant term, the exponent of \(x\) must be zero:

\[ b - c = 0 \implies b = c \]

### Step 3: Substitute \(b = c\) into the equation
Since \(a + b + c = 6\) and \(b = c\), we can substitute \(c\) with \(b\):

\[ a + b + b = 6 \implies a + 2b = 6 \implies a = 6 - 2b \]

### Step 4: Determine possible values for \(b\)
Since \(a\), \(b\), and \(c\) must be non-negative integers, we can find the possible values for \(b\):

- If \(b = 0\), then \(a = 6\) and \(c = 0\).
- If \(b = 1\), then \(a = 4\) and \(c = 1\).
- If \(b = 2\), then \(a = 2\) and \(c = 2\).
- If \(b = 3\), then \(a = 0\) and \(c = 3\).

Thus, \(b\) can take values \(0, 1, 2, 3\).

### Step 5: Calculate the constant term for each valid \(b\)
Now we can calculate the constant term for each case:

1. **For \(b = 0\) (i.e., \(a = 6\), \(c = 0\))**:
\[ T = \frac{6!}{6!0!0!} \cdot 1^6 \cdot 2^0 = 1 \]

2. **For \(b = 1\) (i.e., \(a = 4\), \(c = 1\))**:
\[ T = \frac{6!}{4!1!1!} \cdot 1^4 \cdot 2^1 = \frac{720}{24 \cdot 1 \cdot 1} \cdot 2 = 30 \cdot 2 = 60 \]

3. **For \(b = 2\) (i.e., \(a = 2\), \(c = 2\))**:
\[ T = \frac{6!}{2!2!2!} \cdot 1^2 \cdot 2^2 = \frac{720}{2 \cdot 2 \cdot 2} \cdot 4 = \frac{720}{8} \cdot 4 = 90 \cdot 4 = 360 \]

4. **For \(b = 3\) (i.e., \(a = 0\), \(c = 3\))**:
\[ T = \frac{6!}{0!3!3!} \cdot 1^0 \cdot 2^3 = \frac{720}{1 \cdot 6 \cdot 6} \cdot 8 = \frac{720}{36} \cdot 8 = 20 \cdot 8 = 160 \]

### Step 6: Sum the constant terms
Now, we sum all the constant terms calculated:

\[ 1 + 60 + 360 + 160 = 581 \]

### Final Answer:
The constant term in the expansion of \((1 + x + \frac{2}{x})^6\) is **581**.

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