Find the 7th term from the end in the expansion of `(x+(1)/(x))^(11)`

Explanation

To find the 7th term from the end in the expansion of \( (x + \frac{1}{x})^{11} \), we can follow these steps:

### Step 1: Identify the general term in the binomial expansion
The general term \( T_r \) in the expansion of \( (p + q)^n \) is given by:
\[ T_r = \binom{n}{r-1} p^{n - (r-1)} q^{r-1} \]
In our case, \( p = x \), \( q = \frac{1}{x} \), and \( n = 11 \).

### Step 2: Determine the term from the end
To find the 7th term from the end, we can use the relation:
\[ \text{rth term from the end} = (n - r + 2) \text{th term from the beginning} \]
Here, \( n = 11 \) and we want the 7th term from the end, so:
\[ r = 7 \implies \text{term from the beginning} = 11 - 7 + 2 = 6 \]

### Step 3: Calculate the 6th term from the beginning
Using the general term formula:
\[ T_6 = \binom{11}{6-1} x^{11 - (6-1)} \left(\frac{1}{x}\right)^{6-1} \]
This simplifies to:
\[ T_6 = \binom{11}{5} x^{11 - 5} \left(\frac{1}{x}\right)^{5} \]
\[ = \binom{11}{5} x^6 \cdot \frac{1}{x^5} = \binom{11}{5} x^{6-5} = \binom{11}{5} x^1 \]

### Step 4: Calculate \( \binom{11}{5} \)
Using the formula for combinations:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
We calculate:
\[ \binom{11}{5} = \frac{11!}{5! \cdot 6!} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462 \]

### Step 5: Final term calculation
Thus, the 6th term is:
\[ T_6 = 462 x \]

### Conclusion
Therefore, the 7th term from the end in the expansion of \( (x + \frac{1}{x})^{11} \) is:
\[ \boxed{462x} \]