The number of terms in the expansion of `(1+7sqrt(2x))^9+(1-7sqrt(2x))^9` is

Options

A. 5
B. 7
C. 9
D. 10

Answer

To find the number of terms in the expansion of \((1 + 7\sqrt{2x})^9 + (1 - 7\sqrt{2x})^9\), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Expression**:
The expression consists of two parts: \((1 + 7\sqrt{2x})^9\) and \((1 - 7\sqrt{2x})^9\). We will expand both parts using the Binomial Theorem.

2. **Apply the Binomial Theorem**:
The Binomial Theorem states that:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
For \((1 + 7\sqrt{2x})^9\):
\[ (1 + 7\sqrt{2x})^9 = \sum_{k=0}^{9} \binom{9}{k} (1)^{9-k} (7\sqrt{2x})^k = \sum_{k=0}^{9} \binom{9}{k} 7^k (2x)^{k/2} \]

3. **Consider the Second Part**:
For \((1 - 7\sqrt{2x})^9\):
\[ (1 - 7\sqrt{2x})^9 = \sum_{k=0}^{9} \binom{9}{k} (1)^{9-k} (-7\sqrt{2x})^k = \sum_{k=0}^{9} \binom{9}{k} (-7)^k (2x)^{k/2} \]

4. **Combine the Two Expansions**:
Adding both expansions:
\[ (1 + 7\sqrt{2x})^9 + (1 - 7\sqrt{2x})^9 = \sum_{k=0}^{9} \binom{9}{k} 7^k (2x)^{k/2} + \sum_{k=0}^{9} \binom{9}{k} (-7)^k (2x)^{k/2} \]
Notice that the terms with odd \(k\) will cancel out, while the terms with even \(k\) will double.

5. **Identify the Even Powers**:
The even powers of \(k\) are \(0, 2, 4, 6, 8\). Thus, the relevant terms in the combined expansion are:
- For \(k = 0\): \(x^0\)
- For \(k = 2\): \(x^1\)
- For \(k = 4\): \(x^2\)
- For \(k = 6\): \(x^3\)
- For \(k = 8\): \(x^4\)

6. **Count the Number of Terms**:
The even values of \(k\) from \(0\) to \(8\) are \(0, 2, 4, 6, 8\), which gives us a total of \(5\) terms.

### Final Answer:
The number of terms in the expansion of \((1 + 7\sqrt{2x})^9 + (1 - 7\sqrt{2x})^9\) is **5**.