We are asked to find the middle term in the expansion of $(3x - \frac{x^3}{6})^7$.

Step 1: Identify the binomial expansion formula

The binomial expansion of $(a + b)^n$ is given by:

$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

Here, we identify $a = 3x$, $b = -\frac{x^3}{6}$, and $n = 7$.

Step 2: Find the middle term index

In a binomial expansion of degree $n$, the middle term corresponds to:

• If $n$ is even, the middle term is the $(n/2 + 1)$-th term.
• If $n$ is odd, the middle term is the $((n+1)/2)$-th term.

Since $n = 7$ is odd, the middle term is the $((7+1)/2) = 4$-th term.

Step 3: Apply the binomial expansion formula for the 4th term

The $k$-th term in the binomial expansion is:

$T_{k+1} = \binom{n}{k} a^{n-k} b^k$

For the middle term, $k = 3$:

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

Step 4: Calculate each part

• $\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$
• $(3x)^4 = 3^4 x^4 = 81x^4$
• $\left(-\frac{x^3}{6}\right)^3 = \left(-\frac{1}{6}\right)^3 x^9 = -\frac{1}{216} x^9$

Step 5: Multiply everything together

$T_4 = 35 \times 81x^4 \times \left(-\frac{1}{216} x^9\right)$

Simplify:

$T_4 = 35 \times 81 \times -\frac{1}{216} x^{4+9}$ $T_4 = 35 \times -\frac{81}{216} x^{13}$ $T_4 = 35 \times -\frac{3}{8} x^{13}$ $T_4 = -\frac{105}{8} x^{13}$

Final Answer:

The middle term in the expansion of $(3x - \frac{x^3}{6})^7$ is:

$-\frac{105}{8} x^{13}$

Would you like more details on any of the steps or clarification?

Relative Questions:

1. How does the binomial expansion change for negative powers?
2. Can the binomial theorem be used for non-integer exponents?
3. What are the uses of binomial expansion in calculus?
4. How do you derive the general term in a binomial expansion?
5. What is the importance of combinations in the binomial theorem?

Tip: When solving binomial expansion problems, always check whether the exponent is odd or even to quickly find the middle term.