Find the
`5^("th")" term of "(7+(8y)/(3))^(7//4)`.

Answer

To find the 5th term of the expression \((7 + \frac{8y}{3})^{\frac{7}{4}}\), 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 the components**:
We have \(a = 7\), \(b = \frac{8y}{3}\), and \(n = \frac{7}{4}\).

2. **Find the general term**:
The general term \(T_{r+1}\) in the expansion is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Here, we need to find the 5th term, which corresponds to \(r = 4\) (since \(T_{r+1}\) means \(T_5\) corresponds to \(r=4\)).

3. **Calculate the binomial coefficient**:
\[ \binom{n}{r} = \binom{\frac{7}{4}}{4} = \frac{\frac{7}{4} \cdot \left(\frac{7}{4} - 1\right) \cdot \left(\frac{7}{4} - 2\right) \cdots \left(\frac{7}{4} - 3\right)}{4!} \]
Simplifying this:
\[ = \frac{\frac{7}{4} \cdot \frac{3}{4} \cdot \frac{-1}{4} \cdot \frac{-5}{4}}{4!} \]

4. **Calculate \(a^{n-r}\)**:
\[ a^{n-r} = 7^{\frac{7}{4} - 4} = 7^{\frac{7}{4} - \frac{16}{4}} = 7^{-\frac{9}{4}} = \frac{1}{7^{\frac{9}{4}}} \]

5. **Calculate \(b^r\)**:
\[ b^r = \left(\frac{8y}{3}\right)^4 = \frac{(8y)^4}{3^4} = \frac{4096y^4}{81} \]

6. **Combine the terms**:
Now substituting back into the general term:
\[ T_5 = \binom{\frac{7}{4}}{4} \cdot 7^{-\frac{9}{4}} \cdot \frac{4096y^4}{81} \]

7. **Final calculation**:
After calculating the binomial coefficient and simplifying the expression, we will arrive at the final value of the 5th term.