Find the 7th term in the expansion of `((4x)/5+5/(2x))^(8)`

Answer

To find the 7th term in the expansion of \(\left(\frac{4x}{5} + \frac{5}{2x}\right)^{8}\), we can use the Binomial Theorem. According to the Binomial Theorem, the \(n\)th term in the expansion of \((a + b)^n\) is given by:

\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]

where \(n\) is the power, \(a\) and \(b\) are the terms being expanded, and \(r\) is the term number minus one.

### Step 1: Identify the values
In our case:
- \(n = 8\)
- \(a = \frac{4x}{5}\)
- \(b = \frac{5}{2x}\)

We need to find the 7th term, which corresponds to \(r = 6\) (since \(T_{r+1} = T_7\)).

### Step 2: Substitute into the formula
Using the formula for the \(T_{r+1}\):
\[ T_{7} = \binom{8}{6} \left(\frac{4x}{5}\right)^{8-6} \left(\frac{5}{2x}\right)^{6} \]

### Step 3: Calculate the binomial coefficient
Calculate \(\binom{8}{6}\):
\[ \binom{8}{6} = \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28 \]

### Step 4: Calculate the powers of \(a\) and \(b\)
Now calculate \(\left(\frac{4x}{5}\right)^{2}\) and \(\left(\frac{5}{2x}\right)^{6}\):
\[ \left(\frac{4x}{5}\right)^{2} = \frac{(4x)^{2}}{5^{2}} = \frac{16x^{2}}{25} \]
\[ \left(\frac{5}{2x}\right)^{6} = \frac{5^{6}}{(2x)^{6}} = \frac{15625}{64x^{6}} \]

### Step 5: Combine the terms
Now substitute these back into the expression for \(T_{7}\):
\[ T_{7} = 28 \cdot \frac{16x^{2}}{25} \cdot \frac{15625}{64x^{6}} \]

### Step 6: Simplify the expression
Now simplify:
\[ T_{7} = 28 \cdot \frac{16 \cdot 15625}{25 \cdot 64} \cdot \frac{x^{2}}{x^{6}} = 28 \cdot \frac{16 \cdot 15625}{25 \cdot 64} \cdot \frac{1}{x^{4}} \]

Calculating the constants:
\[ \frac{16}{64} = \frac{1}{4} \quad \text{and} \quad \frac{15625}{25} = 625 \]
Thus,
\[ T_{7} = 28 \cdot \frac{625}{4} \cdot \frac{1}{x^{4}} = \frac{17500}{4x^{4}} = \frac{4375}{x^{4}} \]

### Final Answer
The 7th term in the expansion is:
\[ T_{7} = \frac{4375}{x^{4}} \]