The term independent of x in the expansion of `(1 + x)^n (1 + 1/x)^n` is

Options

A. `C_0^2 + C_1^2 + C_2^2 + ……….+C_n^2`
B. `""^(2n)C_n`
C. `(1.3.5…..(2n - 1))/(n!) 2^n`
D. All of the above

Answer

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

### Step 1: Rewrite the expression
We start with the expression:
\[ (1 + x)^n \left(1 + \frac{1}{x}\right)^n \]
This can be rewritten using the property of exponents as:
\[ (1 + x)^n \cdot \left(\frac{1 + x}{x}\right)^n = \frac{(1 + x)^n (1 + x)^n}{x^n} = \frac{(1 + x)^{2n}}{x^n} \]

### Step 2: Identify the general term
The general term in the expansion of \( (1 + x)^{2n} \) is given by:
\[ T_{r+1} = \binom{2n}{r} x^r \]
Thus, the general term of the entire expression becomes:
\[ T_{r+1} = \frac{\binom{2n}{r} x^r}{x^n} = \binom{2n}{r} x^{r-n} \]

### Step 3: Set the exponent of \( x \) to zero
To find the term independent of \( x \), we need to set the exponent of \( x \) to zero:
\[ r - n = 0 \implies r = n \]

### Step 4: Substitute \( r \) back into the general term
Now, substituting \( r = n \) back into the general term:
\[ T_{n+1} = \binom{2n}{n} x^{n-n} = \binom{2n}{n} x^0 = \binom{2n}{n} \]

### Conclusion
Thus, the term independent of \( x \) in the expansion of \( (1 + x)^n (1 + \frac{1}{x})^n \) is:
\[ \boxed{\binom{2n}{n}} \]

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