Find the middle term in the expansion of `(2x^2-1/x)^7`

Answer

To find the middle term in the expansion of \((2x^2 - \frac{1}{x})^7\), we can follow these steps:

### Step 1: Identify the value of \(n\)
The expression is raised to the power of \(7\), so we have:
\[ n = 7 \]

### Step 2: Determine the number of terms
The total number of terms in the expansion of \((a + b)^n\) is given by \(n + 1\). Therefore, the number of terms is:
\[ n + 1 = 7 + 1 = 8 \]

### Step 3: Identify the middle terms
Since the total number of terms is \(8\) (which is even), there will be two middle terms. The middle terms are the \(4^{th}\) and \(5^{th}\) terms.

### Step 4: Use the Binomial Theorem
The \(r^{th}\) term in the expansion of \((a + b)^n\) is given by:
\[ T_{r} = \binom{n}{r-1} a^{n-(r-1)} b^{r-1} \]
For our case, \(a = 2x^2\) and \(b = -\frac{1}{x}\).

### Step 5: Calculate the \(4^{th}\) term
For the \(4^{th}\) term (\(r = 4\)):
\[ T_{4} = \binom{7}{3} (2x^2)^{7-3} \left(-\frac{1}{x}\right)^{3} \]
Calculating each part:
- \(\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35\)
- \((2x^2)^{4} = 2^4 (x^2)^4 = 16x^8\)
- \(\left(-\frac{1}{x}\right)^{3} = -\frac{1}{x^3}\)

Putting it all together:
\[ T_{4} = 35 \cdot 16x^8 \cdot \left(-\frac{1}{x^3}\right) = -35 \cdot 16 \cdot x^{8-3} = -560x^5 \]

### Step 6: Calculate the \(5^{th}\) term
For the \(5^{th}\) term (\(r = 5\)):
\[ T_{5} = \binom{7}{4} (2x^2)^{7-4} \left(-\frac{1}{x}\right)^{4} \]
Calculating each part:
- \(\binom{7}{4} = \binom{7}{3} = 35\)
- \((2x^2)^{3} = 2^3 (x^2)^3 = 8x^6\)
- \(\left(-\frac{1}{x}\right)^{4} = \frac{1}{x^4}\)

Putting it all together:
\[ T_{5} = 35 \cdot 8x^6 \cdot \frac{1}{x^4} = 35 \cdot 8 \cdot x^{6-4} = 280x^2 \]

### Final Result
The middle terms in the expansion of \((2x^2 - \frac{1}{x})^7\) are:
\[ T_{4} = -560x^5 \quad \text{and} \quad T_{5} = 280x^2 \]