Expand using binomial theorem `(x+1/x)^2 (x ne 0)` = ………..

Answer

To expand the expression \((x + \frac{1}{x})^2\) using the binomial theorem, we can follow these steps:

### Step 1: Identify the terms
In the expression \((x + \frac{1}{x})^2\), we can identify:
- \(a = x\)
- \(b = \frac{1}{x}\)
- \(n = 2\)

### Step 2: Apply the binomial theorem
The binomial theorem states that:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
For our case, we substitute \(a\), \(b\), and \(n\):
\[ (x + \frac{1}{x})^2 = \sum_{k=0}^{2} \binom{2}{k} x^{2-k} \left(\frac{1}{x}\right)^k \]

### Step 3: Calculate each term in the expansion
Now we calculate each term for \(k = 0\), \(1\), and \(2\):

- **For \(k = 0\)**:
\[ \binom{2}{0} x^{2-0} \left(\frac{1}{x}\right)^0 = 1 \cdot x^2 \cdot 1 = x^2 \]

- **For \(k = 1\)**:
\[ \binom{2}{1} x^{2-1} \left(\frac{1}{x}\right)^1 = 2 \cdot x^1 \cdot \frac{1}{x} = 2 \]

- **For \(k = 2\)**:
\[ \binom{2}{2} x^{2-2} \left(\frac{1}{x}\right)^2 = 1 \cdot x^0 \cdot \frac{1}{x^2} = \frac{1}{x^2} \]

### Step 4: Combine all terms
Now, we combine all the terms from \(k = 0\), \(1\), and \(2\):
\[ (x + \frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2} \]

### Final Result
Thus, the expansion of \((x + \frac{1}{x})^2\) is:
\[ x^2 + 2 + \frac{1}{x^2} \]

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