Find the middle term in the expansion of `(3-(x^(3))/(6))^(7)`

Explanation

To find the middle term in the expansion of \( (3 - \frac{x^3}{6})^7 \), we will use the Binomial Theorem. The Binomial Theorem states that:

\[ (p + q)^n = \sum_{r=0}^{n} \binom{n}{r} p^{n-r} q^r \]

### Step 1: Identify \( p \), \( q \), and \( n \)

In our case, we have:
- \( p = 3 \)
- \( q = -\frac{x^3}{6} \)
- \( n = 7 \)

### Step 2: Determine the general term

The general term \( T_{r+1} \) in the expansion is given by:

\[ T_{r+1} = \binom{n}{r} p^{n-r} q^r \]

Substituting our values, we get:

\[ T_{r+1} = \binom{7}{r} (3)^{7-r} \left(-\frac{x^3}{6}\right)^r \]

### Step 3: Simplify the general term

This simplifies to:

\[ T_{r+1} = \binom{7}{r} (3)^{7-r} \left(-1\right)^r \left(\frac{x^3}{6}\right)^r \]

\[ = \binom{7}{r} (3)^{7-r} \left(-1\right)^r \frac{x^{3r}}{6^r} \]

\[ = \binom{7}{r} (3)^{7-r} (-1)^r \frac{x^{3r}}{6^r} \]

### Step 4: Find the middle term

Since \( n = 7 \) (an odd number), the middle terms are \( T_{4} \) and \( T_{5} \) (the 4th and 5th terms).

### Step 5: Calculate \( T_4 \)

For \( T_4 \) (when \( r = 3 \)):

\[ T_4 = \binom{7}{3} (3)^{7-3} \left(-\frac{x^3}{6}\right)^3 \]

Calculating each part:

1. \( \binom{7}{3} = 35 \)
2. \( (3)^{4} = 81 \)
3. \( \left(-\frac{x^3}{6}\right)^3 = -\frac{x^9}{216} \)

Putting it all together:

\[ T_4 = 35 \cdot 81 \cdot \left(-\frac{x^9}{216}\right) = -\frac{2835 x^9}{216} \]

### Step 6: Calculate \( T_5 \)

For \( T_5 \) (when \( r = 4 \)):

\[ T_5 = \binom{7}{4} (3)^{7-4} \left(-\frac{x^3}{6}\right)^4 \]

Calculating each part:

1. \( \binom{7}{4} = 35 \)
2. \( (3)^{3} = 27 \)
3. \( \left(-\frac{x^3}{6}\right)^4 = \frac{x^{12}}{1296} \)

Putting it all together:

\[ T_5 = 35 \cdot 27 \cdot \frac{x^{12}}{1296} = \frac{945 x^{12}}{1296} \]

### Final Result

The middle terms in the expansion of \( (3 - \frac{x^3}{6})^7 \) are:

1. \( T_4 = -\frac{2835 x^9}{216} \)
2. \( T_5 = \frac{945 x^{12}}{1296} \)