For ` |x|>2/3` ​ ,find the value of the third term in the expansion of `(2x+3)^(3/5)`

Answer

To find the third term in the expansion of \((2x + 3)^{\frac{3}{5}}\) for \(|x| > \frac{2}{3}\), we can use the Binomial Theorem. The Binomial Theorem states that:

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

where \(\binom{n}{k}\) is the binomial coefficient.

### Step-by-Step Solution:

1. **Identify \(a\), \(b\), and \(n\)**:
- Here, \(a = 2x\), \(b = 3\), and \(n = \frac{3}{5}\).

2. **Write the general term**:
- The \(k\)-th term in the expansion is given by:
\[ T_k = \binom{n}{k} a^{n-k} b^k \]
- For our case, it becomes:
\[ T_k = \binom{\frac{3}{5}}{k} (2x)^{\frac{3}{5} - k} (3)^k \]

3. **Find the third term**:
- The third term corresponds to \(k = 2\) (since \(k\) starts from 0).
\[ T_2 = \binom{\frac{3}{5}}{2} (2x)^{\frac{3}{5} - 2} (3)^2 \]

4. **Calculate the binomial coefficient**:
- The binomial coefficient \(\binom{\frac{3}{5}}{2}\) is calculated as:
\[ \binom{\frac{3}{5}}{2} = \frac{\frac{3}{5} \left(\frac{3}{5} - 1\right)}{2!} = \frac{\frac{3}{5} \cdot \left(-\frac{2}{5}\right)}{2} = -\frac{3 \cdot 2}{5 \cdot 5 \cdot 2} = -\frac{3}{50} \]

5. **Substitute values into the term**:
- Now substituting back into the term:
\[ T_2 = -\frac{3}{50} (2x)^{\frac{3}{5} - 2} (3)^2 \]
- Simplifying \((2x)^{\frac{3}{5} - 2}\):
\[ (2x)^{\frac{3}{5} - 2} = (2x)^{\frac{3}{5} - \frac{10}{5}} = (2x)^{-\frac{7}{5}} = \frac{1}{(2x)^{\frac{7}{5}}} \]

6. **Combine everything**:
- Thus, we have:
\[ T_2 = -\frac{3}{50} \cdot \frac{9}{(2x)^{\frac{7}{5}}} = -\frac{27}{50(2x)^{\frac{7}{5}}} \]

7. **Final expression**:
- Therefore, the third term in the expansion of \((2x + 3)^{\frac{3}{5}}\) is:
\[ T_2 = -\frac{27}{100} x^{-\frac{7}{5}} \]

### Final Answer:
The third term in the expansion of \((2x + 3)^{\frac{3}{5}}\) is:
\[ -\frac{27}{100} x^{-\frac{7}{5}} \]