If `(1+x-2x^2)^8 = 1 + a_1x + a_2x^2 + ……+ a_16 x^16`, then `a_1 +a_3 + a_5 + ……+a_15 = ?`

Options

A. `2^7`
B. `-2^7`
C. `3^2`
D. `4^6`

Answer

To solve the problem, we need to find the sum \( a_1 + a_3 + a_5 + \ldots + a_{15} \) from the expansion of \( (1 + x - 2x^2)^8 \).

### Step-by-Step Solution:

1. **Write the expression**:
We start with the expression given in the problem:
\[ (1 + x - 2x^2)^8 = 1 + a_1 x + a_2 x^2 + \ldots + a_{16} x^{16} \]

2. **Substitute \( -x \)**:
To find the sum of the coefficients of the odd powers of \( x \), we can substitute \( -x \) into the expression:
\[ (1 - x - 2(-x^2))^8 = (1 - x + 2x^2)^8 \]
This gives us:
\[ 1 - a_1 x + a_2 x^2 - a_3 x^3 + a_4 x^4 - a_5 x^5 + \ldots - a_{16} x^{16} \]

3. **Set up the equation**:
Now we have two equations:
\[ (1 + x - 2x^2)^8 = 1 + a_1 x + a_2 x^2 + \ldots + a_{16} x^{16} \tag{1} \]
\[ (1 - x + 2x^2)^8 = 1 - a_1 x + a_2 x^2 - a_3 x^3 + \ldots - a_{16} x^{16} \tag{2} \]

4. **Subtract the two equations**:
Subtract equation (2) from equation (1):
\[ (1 + x - 2x^2)^8 - (1 - x + 2x^2)^8 = (a_1 + a_3 + a_5 + \ldots + a_{15}) \cdot 2x \]
The even powers will cancel out, leaving only the odd powers.

5. **Evaluate at \( x = 1 \)**:
Now, we evaluate both sides at \( x = 1 \):
\[ (1 + 1 - 2 \cdot 1^2)^8 - (1 - 1 + 2 \cdot 1^2)^8 = 2(a_1 + a_3 + a_5 + \ldots + a_{15}) \]
Simplifying this gives:
\[ (1 + 1 - 2)^8 - (1 - 1 + 2)^8 = 2(a_1 + a_3 + a_5 + \ldots + a_{15}) \]
\[ (0)^8 - (2)^8 = 2(a_1 + a_3 + a_5 + \ldots + a_{15}) \]
\[ 0 - 256 = 2(a_1 + a_3 + a_5 + \ldots + a_{15}) \]
\[ -256 = 2(a_1 + a_3 + a_5 + \ldots + a_{15}) \]

6. **Solve for the sum**:
Dividing both sides by 2 gives:
\[ a_1 + a_3 + a_5 + \ldots + a_{15} = -128 \]

### Final Answer:
\[ a_1 + a_3 + a_5 + \ldots + a_{15} = -128 \]