No. of terms in the expansion of `(1+2x)^(9) +(1-2x)^(9)` is :

Options

A. 10
B. 9
C. 7
D. 20

Explanation

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

### Step 1: Expand both expressions using the Binomial Theorem.
Using the Binomial Theorem, the expansion of \((a + b)^n\) is given by:

\[ T_k = \binom{n}{k} a^{n-k} b^k \]

For \((1 + 2x)^{9}\), the general term is:

\[ T_k = \binom{9}{k} (1)^{9-k} (2x)^k = \binom{9}{k} 2^k x^k \]

For \((1 - 2x)^{9}\), the general term is:

\[ T_k = \binom{9}{k} (1)^{9-k} (-2x)^k = \binom{9}{k} (-2)^k x^k \]

### Step 2: Combine the expansions.
Now, we combine the two expansions:

\[ (1 + 2x)^{9} + (1 - 2x)^{9} = \sum_{k=0}^{9} \left( \binom{9}{k} 2^k x^k + \binom{9}{k} (-2)^k x^k \right) \]

### Step 3: Simplify the combined terms.
Notice that:

- If \(k\) is even, \( (-2)^k = 2^k \), so the terms will add up.
- If \(k\) is odd, \( (-2)^k = -2^k \), so the terms will cancel out.

Thus, the combined expression simplifies to:

\[ \sum_{k \text{ even}} \binom{9}{k} 2^k x^k \]

### Step 4: Determine the values of \(k\).
The even values of \(k\) from \(0\) to \(9\) are \(0, 2, 4, 6, 8\).

### Step 5: Count the number of terms.
The even integers from \(0\) to \(9\) are:

- \(0\)
- \(2\)
- \(4\)
- \(6\)
- \(8\)

This gives us a total of \(5\) terms.

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

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