Find the coefficient of
(ii) `x^7` in the expansion of `(x^(2) + (1)/(x) )^(11)`

Answer

To find the coefficient of \( x^7 \) in the expansion of \( (x^2 + \frac{1}{x})^{11} \), we can use the Binomial Theorem.

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term in the binomial expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \( a = x^2 \), \( b = \frac{1}{x} \), and \( n = 11 \). Thus, the general term becomes:
\[ T_{r+1} = \binom{11}{r} (x^2)^{11-r} \left(\frac{1}{x}\right)^r \]

2. **Simplify the General Term**:
Simplifying \( T_{r+1} \):
\[ T_{r+1} = \binom{11}{r} x^{2(11-r)} \cdot x^{-r} = \binom{11}{r} x^{22 - 2r - r} = \binom{11}{r} x^{22 - 3r} \]

3. **Set the Power of \( x \) Equal to 7**:
We need to find the term where the power of \( x \) is 7:
\[ 22 - 3r = 7 \]
Solving for \( r \):
\[ 22 - 7 = 3r \implies 15 = 3r \implies r = 5 \]

4. **Substitute \( r \) Back into the General Term**:
Now we substitute \( r = 5 \) back into the general term to find the coefficient:
\[ T_{6} = \binom{11}{5} x^{22 - 3 \cdot 5} = \binom{11}{5} x^{7} \]

5. **Calculate the Coefficient**:
We need to calculate \( \binom{11}{5} \):
\[ \binom{11}{5} = \frac{11!}{5!(11-5)!} = \frac{11!}{5! \cdot 6!} \]
Expanding this:
\[ \binom{11}{5} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = \frac{55440}{120} = 462 \]

6. **Final Answer**:
Therefore, the coefficient of \( x^7 \) in the expansion of \( (x^2 + \frac{1}{x})^{11} \) is \( 462 \).