Find the number of terms in the expansion of `(4x - 7y)^49 + (4x + 7y)^49 `

Answer

To find the number of terms in the expansion of \( (4x - 7y)^{49} + (4x + 7y)^{49} \), we can follow these steps:

### Step 1: Understanding the Binomial Expansion
The binomial expansion of \( (a + b)^n \) is given by:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
In our case, we have two expressions: \( (4x - 7y)^{49} \) and \( (4x + 7y)^{49} \).

### Step 2: Expanding Both Expressions
1. **Expansion of \( (4x - 7y)^{49} \)**:
\[ (4x - 7y)^{49} = \sum_{k=0}^{49} \binom{49}{k} (4x)^{49-k} (-7y)^k \]

2. **Expansion of \( (4x + 7y)^{49} \)**:
\[ (4x + 7y)^{49} = \sum_{k=0}^{49} \binom{49}{k} (4x)^{49-k} (7y)^k \]

### Step 3: Combining the Two Expansions
Now, we add both expansions:
\[ (4x - 7y)^{49} + (4x + 7y)^{49} = \sum_{k=0}^{49} \binom{49}{k} (4x)^{49-k} \left[ (-7y)^k + (7y)^k \right] \]

### Step 4: Simplifying the Expression
Notice that:
- For even \( k \), \( (-7y)^k + (7y)^k = 2(7y)^k \)
- For odd \( k \), \( (-7y)^k + (7y)^k = 0 \)

Thus, only the terms where \( k \) is even will contribute to the final sum.

### Step 5: Counting the Number of Terms
The values of \( k \) can be \( 0, 2, 4, \ldots, 48 \). This is an arithmetic sequence where:
- The first term \( a = 0 \)
- The last term \( l = 48 \)
- The common difference \( d = 2 \)

To find the number of terms \( n \) in this sequence, we can use the formula for the \( n \)-th term of an arithmetic sequence:
\[ l = a + (n-1)d \]
Substituting the known values:
\[ 48 = 0 + (n-1) \cdot 2 \]
\[ 48 = (n-1) \cdot 2 \]
\[ n-1 = 24 \quad \Rightarrow \quad n = 25 \]

### Final Answer
Thus, the number of terms in the expansion of \( (4x - 7y)^{49} + (4x + 7y)^{49} \) is **25**.

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