Write down and simplify
`r^("th")" term in "((3a)/(5)+(5b)/(7))^(8)" "(1le r le9)`

Answer

To find the \( r^{th} \) term in the expansion of \( \left( \frac{3a}{5} + \frac{5b}{7} \right)^{8} \), we will use the Binomial Theorem.

### Step-by-Step Solution:

1. **Identify the parameters**:
The expression is \( \left( \frac{3a}{5} + \frac{5b}{7} \right)^{8} \).
Here, \( n = 8 \), \( x = \frac{3a}{5} \), and \( y = \frac{5b}{7} \).

2. **Use the Binomial Theorem**:
The \( r^{th} \) term (where \( r \) is the term number starting from 0) in the expansion of \( (x + y)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} x^{n-r} y^{r} \]
For our case, we need to find the \( r^{th} \) term, which corresponds to \( T_{r} \):
\[ T_{r} = \binom{n}{r-1} x^{n-(r-1)} y^{r-1} \]

3. **Substitute the values**:
Substitute \( n = 8 \), \( x = \frac{3a}{5} \), and \( y = \frac{5b}{7} \):
\[ T_{r} = \binom{8}{r-1} \left(\frac{3a}{5}\right)^{8-(r-1)} \left(\frac{5b}{7}\right)^{r-1} \]

4. **Simplify the expression**:
This simplifies to:
\[ T_{r} = \binom{8}{r-1} \left(\frac{3a}{5}\right)^{9-r} \left(\frac{5b}{7}\right)^{r-1} \]

5. **Further simplification**:
Now, we can express it as:
\[ T_{r} = \binom{8}{r-1} \left(\frac{3^{9-r} a^{9-r}}{5^{9-r}}\right) \left(\frac{5^{r-1} b^{r-1}}{7^{r-1}}\right) \]
Combining the terms gives:
\[ T_{r} = \binom{8}{r-1} \cdot \frac{3^{9-r} a^{9-r} \cdot 5^{r-1} b^{r-1}}{5^{9-r} \cdot 7^{r-1}} \]
This can be simplified to:
\[ T_{r} = \binom{8}{r-1} \cdot \frac{3^{9-r} a^{9-r} \cdot b^{r-1}}{5^{8} \cdot 7^{r-1}} \]

### Final Expression:
Thus, the \( r^{th} \) term in the expansion of \( \left( \frac{3a}{5} + \frac{5b}{7} \right)^{8} \) is:
\[ T_{r} = \binom{8}{r-1} \cdot \frac{3^{9-r} a^{9-r} \cdot b^{r-1}}{5^{8} \cdot 7^{r-1}} \]