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The question asks to find the middle terms in the expansion of the following three binomials:

(i) \((x - \frac{1}{x})^{15}\)

(ii) \((3a - \frac{x^3}{6})^9\)

(iii) \((2x^2 + \frac{y}{x})^9\)

### Approach:
To solve these, we'll use the **Binomial Theorem**, which states that:

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

For each case, we'll find the middle term by determining the general term \( T_{r+1} \) from the binomial expansion, which is given by:

\[
T_{r+1} = \binom{n}{r} a^{n-r} b^r
\]

Then, we calculate for the specific middle term by finding the appropriate value of \( r \).

Let’s break down each part:

---

### (i) \((x - \frac{1}{x})^{15}\):

Here, \( a = x \) and \( b = -\frac{1}{x} \).

- The number of terms is \( n+1 = 16 \), which is even.
- So, the middle term will be the 8th term (since the middle of 16 terms is the 8th and 9th terms).

Let the general term be:
\[
T_{r+1} = \binom{15}{r} x^{15-r} \left( -\frac{1}{x} \right)^r
\]

Simplifying:
\[
T_{r+1} = \binom{15}{r} (-1)^r x^{15-2r}
\]

The middle terms correspond to \( r = 7 \) (and possibly \( r = 8 \) for an even \( n \)).

---

### (ii) \((3a - \frac{x^3}{6})^9\):

Here, \( a = 3a \) and \( b = -\frac{x^3}{6} \).

- The number of terms is \( n+1 = 10 \), so the middle term is the 5th term.

The general term is:
\[
T_{r+1} = \binom{9}{r} (3a)^{9-r} \left( -\frac{x^3}{6} \right)^r
\]

Simplifying:
\[
T_{r+1} = \binom{9}{r} (-1)^r 3^{9-r} a^{9-r} \left( \frac{x^3}{6} \right)^r
\]

The middle term corresponds to \( r = 4 \).

---

### (iii) \((2x^2 + \frac{y}{x})^9\):

Here, \( a = 2x^2 \) and \( b = \frac{y}{x} \).

- The number of terms is \( n+1 = 10 \), so the middle term is the 5th term.

The general term is:
\[
T_{r+1} = \binom{9}{r} (2x^2)^{9-r} \left( \frac{y}{x} \right)^r
\]

Simplifying:
\[
T_{r+1} = \binom{9}{r} 2^{9-r} x^{2(9-r)} \cdot y^r \cdot x^{-r}
\]

This becomes:
\[
T_{r+1} = \binom{9}{r} 2^{9-r} y^r x^{18-3r}
\]

The middle term corresponds to \( r = 4 \).

---

Would you like more detailed steps or further assistance on any specific part?

### 5 Related Questions:
1. How do we determine the middle term if \( n \) is odd?
2. What happens if the binomial has negative powers?
3. Can you explain the role of \( (-1)^r \) in binomial expansions?
4. How would the solution change for a binomial like \( (a + b)^n \) where both terms are negative?
5. How do we calculate combinations \( \binom{n}{r} \) quickly for large \( n \)?

### Tip:
Always simplify the terms in a binomial expansion step by step to avoid errors with negative signs or exponents.

Find the middle terms in the expansion of (i) (x - 1/x)^15 (ii) (3a - x^3/6)^9 (iii) (2x^2 + y/x)^9

Solve and find the middle terms for the binomial expansions of (x - 1/x)^15, (3a - x^3/6)^9, and (2x^2 + y/x)^9. Learn how to apply the binomial theorem to find specific terms in the expansion. Suitable for advanced algebra students in Grades 11-12.

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