If `(1+x -2x^2)^8 =1 +a_1x + a_2x^2+……+a_16x^16, ` then `a_2+a_4+a_6+……+a_16` = ?

Answer

To solve the problem, we need to find the sum of the coefficients \( a_2 + a_4 + a_6 + \ldots + a_{16} \) from the expansion of \( (1 + x - 2x^2)^8 \).

### Step-by-Step Solution:

1. **Identify the expression to expand**:
We are given the expression \( (1 + x - 2x^2)^8 \).

2. **Use the Binomial Theorem**:
According to the Binomial Theorem, the expansion of \( (a + b + c)^n \) can be expressed as a sum of terms of the form \( \frac{n!}{p!q!r!} a^p b^q c^r \) where \( p + q + r = n \).

3. **Set up the equations**:
We want to find the coefficients of even powers of \( x \). To do this, we will evaluate the expression at \( x = 1 \) and \( x = -1 \).

4. **Evaluate at \( x = 1 \)**:
\[ (1 + 1 - 2 \cdot 1^2)^8 = (1 + 1 - 2)^8 = 0^8 = 0 \]
This gives us the equation:
\[ 0 = 1 + a_1 + a_2 + a_3 + \ldots + a_{16} \quad \text{(Equation 1)} \]

5. **Evaluate at \( x = -1 \)**:
\[ (1 - 1 - 2 \cdot (-1)^2)^8 = (1 - 1 - 2)^8 = (-2)^8 = 256 \]
This gives us the equation:
\[ 256 = 1 - a_1 + a_2 - a_3 + \ldots + a_{16} \quad \text{(Equation 2)} \]

6. **Add the two equations**:
Adding Equation 1 and Equation 2:
\[ 0 + 256 = (1 + a_1 + a_2 + a_3 + \ldots + a_{16}) + (1 - a_1 + a_2 - a_3 + \ldots + a_{16}) \]
This simplifies to:
\[ 256 = 2 + 2(a_2 + a_4 + a_6 + \ldots + a_{16}) \]

7. **Solve for the sum of coefficients**:
Rearranging gives:
\[ 256 - 2 = 2(a_2 + a_4 + a_6 + \ldots + a_{16}) \]
\[ 254 = 2(a_2 + a_4 + a_6 + \ldots + a_{16}) \]
Dividing by 2:
\[ a_2 + a_4 + a_6 + \ldots + a_{16} = 127 \]

### Final Answer:
\[ a_2 + a_4 + a_6 + \ldots + a_{16} = 127 \]