What is the number of non zero terms in the expansion of `( 1+ 2 sqrt( 3x))^(11) + ( 1- 2 sqrt( 3x))^(11)` ?

Options

A. 4
B. 5
C. 6
D. 11

Answer

To find the number of non-zero terms in the expansion of \( (1 + 2\sqrt{3x})^{11} + (1 - 2\sqrt{3x})^{11} \), we can use the properties of binomial expansions.

### Step-by-step Solution:

1. **Identify the structure of the expression**:
The expression can be rewritten as \( (a + b)^n + (a - b)^n \) where \( a = 1 \), \( b = 2\sqrt{3x} \), and \( n = 11 \).

2. **Apply the Binomial Theorem**:
According to the Binomial Theorem, the expansion of \( (a + b)^n \) is given by:
\[ \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
For our case, the expansions will yield terms with both even and odd powers of \( b \).

3. **Combine the expansions**:
The terms from \( (1 + 2\sqrt{3x})^{11} \) will include both odd and even powers of \( \sqrt{3x} \), while those from \( (1 - 2\sqrt{3x})^{11} \) will cancel out the odd powers due to the negative sign.

4. **Determine the number of non-zero terms**:
The non-zero terms in the final expression will only be the even powers of \( b \) from the expansions.

5. **Count the even powers**:
The powers of \( b \) in the expansion are \( 0, 2, 4, 6, 8, 10 \) (since \( b = 2\sqrt{3x} \) and \( n = 11 \) is odd). The even powers range from \( 0 \) to \( 10 \), which gives us:
- \( k = 0 \) (term: \( 1 \))
- \( k = 2 \) (term: \( (2\sqrt{3x})^2 \))
- \( k = 4 \) (term: \( (2\sqrt{3x})^4 \))
- \( k = 6 \) (term: \( (2\sqrt{3x})^6 \))
- \( k = 8 \) (term: \( (2\sqrt{3x})^8 \))
- \( k = 10 \) (term: \( (2\sqrt{3x})^{10} \))

This gives us a total of \( 6 \) non-zero terms.

6. **Final Answer**:
Therefore, the number of non-zero terms in the expansion is \( 6 \).