`15^(th)` term of `(2x-3y)^(20)`

Answer

To find the 15th term of the expression \((2x - 3y)^{20}\), we can use the binomial theorem, which states that:

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

In this case, we have:
- \(a = 2x\)
- \(b = -3y\)
- \(n = 20\)

### Step-by-Step Solution:

1. **Identify the term to find**: We want to find the 15th term of the expansion. In the binomial expansion, the \(r\)-th term is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Since we want the 15th term, we set \(r = 14\) (because the first term corresponds to \(r = 0\)).

2. **Substitute the values**:
\[ T_{15} = \binom{20}{14} (2x)^{20-14} (-3y)^{14} \]

3. **Calculate the binomial coefficient**:
\[ \binom{20}{14} = \binom{20}{6} = \frac{20!}{6!(20-6)!} = \frac{20!}{6! \cdot 14!} \]

4. **Calculate the powers**:
- \( (2x)^{20-14} = (2x)^6 = 2^6 x^6 \)
- \( (-3y)^{14} = (-3)^{14} y^{14} = 3^{14} y^{14} \) (since \((-3)^{14}\) is positive)

5. **Combine the results**:
\[ T_{15} = \binom{20}{14} \cdot 2^6 \cdot x^6 \cdot 3^{14} \cdot y^{14} \]

6. **Final expression**:
\[ T_{15} = \binom{20}{14} \cdot 64 \cdot x^6 \cdot 3^{14} \cdot y^{14} \]

### Final Answer:
\[ T_{15} = \binom{20}{14} \cdot 64 \cdot 3^{14} \cdot x^6 \cdot y^{14} \]