Find the tenth term in the expansion `(2x-y)^(11)`.

Answer

To find the tenth term in the expansion of \((2x - y)^{11}\), we can use the Binomial Theorem. According to the theorem, the \(r\)th term in the expansion of \((a + b)^n\) is given by:

\[ T_r = \binom{n}{r-1} a^{r-1} b^{n - (r-1)} \]

In our case, we have \(a = 2x\), \(b = -y\), and \(n = 11\). We want to find the 10th term, which corresponds to \(r = 10\).

### Step 1: Identify the values
- \(n = 11\)
- \(r = 10\)
- \(a = 2x\)
- \(b = -y\)

### Step 2: Substitute into the formula
Using the formula for the \(r\)th term:

\[ T_{10} = \binom{11}{10-1} (2x)^{10-1} (-y)^{11 - (10-1)} \]

This simplifies to:

\[ T_{10} = \binom{11}{9} (2x)^{9} (-y)^{2} \]

### Step 3: Calculate \(\binom{11}{9}\)
\[ \binom{11}{9} = \binom{11}{2} = \frac{11!}{9! \cdot 2!} = \frac{11 \times 10}{2 \times 1} = 55 \]

### Step 4: Calculate \((2x)^9\)
\[ (2x)^9 = 2^9 \cdot x^9 = 512x^9 \]

### Step 5: Calculate \((-y)^2\)
\[ (-y)^2 = y^2 \]

### Step 6: Combine all parts
Now substituting everything back into the term:

\[ T_{10} = 55 \cdot 512x^9 \cdot y^2 \]

### Step 7: Final calculation
Calculating \(55 \cdot 512\):

\[ T_{10} = 28160x^9y^2 \]

Thus, the tenth term in the expansion of \((2x - y)^{11}\) is:

\[ \boxed{28160x^9y^2} \]