Find the fourth term in the expansion of ` (4/7 x - y^2)^5`

Answer

To find the fourth term in the expansion of \( \left( \frac{4}{7} x - y^2 \right)^5 \), we can use the Binomial Theorem. The Binomial Theorem states that:

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

In our case, we have:

- \( a = \frac{4}{7} x \)
- \( b = -y^2 \)
- \( n = 5 \)

### Step 1: Identify the term we need
We need to find the fourth term, which corresponds to \( T_4 \). According to the Binomial Theorem, the \( (r+1) \)-th term is given by:

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

For the fourth term, \( r + 1 = 4 \) implies \( r = 3 \).

### Step 2: Substitute values into the formula
Now we can substitute \( n = 5 \), \( r = 3 \), \( a = \frac{4}{7} x \), and \( b = -y^2 \) into the formula:

\[ T_4 = \binom{5}{3} \left( \frac{4}{7} x \right)^{5-3} (-y^2)^3 \]

### Step 3: Calculate the binomial coefficient
Calculate \( \binom{5}{3} \):

\[ \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10 \]

### Step 4: Calculate the powers
Now calculate the powers:

1. \( \left( \frac{4}{7} x \right)^{2} = \left( \frac{4}{7} \right)^{2} x^{2} = \frac{16}{49} x^{2} \)
2. \( (-y^2)^{3} = -y^{6} \)

### Step 5: Combine all parts
Now combine everything:

\[ T_4 = 10 \cdot \frac{16}{49} x^{2} \cdot (-y^{6}) = -\frac{160}{49} x^{2} y^{6} \]

### Final Answer
Thus, the fourth term in the expansion is:

\[ T_4 = -\frac{160}{49} x^{2} y^{6} \]