In the expansion of `(x^(2) + (1)/(x) )^(n)`, the coefficient of the fourth term is equal to the coefficient of the ninth term. Find `n` and the sixth term of the expansion.

Answer

To solve the problem, we need to find the value of \( n \) such that the coefficient of the fourth term in the expansion of \( (x^2 + \frac{1}{x})^n \) is equal to the coefficient of the ninth term. We will also find the sixth term of the expansion.

### Step 1: Identify the general term in the binomial expansion

The general term \( T_{r+1} \) in the expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \( a = x^2 \) and \( b = \frac{1}{x} \). Therefore, the general term becomes:
\[ T_{r+1} = \binom{n}{r} (x^2)^{n-r} \left(\frac{1}{x}\right)^r = \binom{n}{r} x^{2(n-r)} x^{-r} = \binom{n}{r} x^{2n - 2r - r} = \binom{n}{r} x^{2n - 3r} \]

### Step 2: Find the coefficients of the fourth and ninth terms

The fourth term corresponds to \( r = 3 \):
\[ T_4 = T_{3+1} = \binom{n}{3} x^{2n - 3 \cdot 3} = \binom{n}{3} x^{2n - 9} \]
The coefficient of the fourth term is \( \binom{n}{3} \).

The ninth term corresponds to \( r = 8 \):
\[ T_9 = T_{8+1} = \binom{n}{8} x^{2n - 3 \cdot 8} = \binom{n}{8} x^{2n - 24} \]
The coefficient of the ninth term is \( \binom{n}{8} \).

### Step 3: Set the coefficients equal

According to the problem, the coefficients of the fourth and ninth terms are equal:
\[ \binom{n}{3} = \binom{n}{8} \]

### Step 4: Use the property of binomial coefficients

The equality \( \binom{n}{k} = \binom{n}{n-k} \) gives us:
\[ \binom{n}{3} = \binom{n}{8} \implies n - 3 = 8 \implies n = 11 \]

### Step 5: Find the sixth term

Now that we have \( n = 11 \), we can find the sixth term \( T_6 \):
\[ T_6 = T_{5+1} = \binom{11}{5} (x^2)^{11-5} \left(\frac{1}{x}\right)^5 = \binom{11}{5} x^{2 \cdot 6} x^{-5} = \binom{11}{5} x^{12 - 5} = \binom{11}{5} x^7 \]

### Step 6: Calculate \( \binom{11}{5} \)

To calculate \( \binom{11}{5} \):
\[ \binom{11}{5} = \frac{11!}{5!(11-5)!} = \frac{11!}{5!6!} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462 \]

### Final Result

Thus, the sixth term is:
\[ T_6 = 462 x^7 \]

### Summary

- The value of \( n \) is \( 11 \).
- The sixth term of the expansion is \( 462 x^7 \).