If `(1 + x + x^(2))^(48) = a_(0) + a_(1) x + a_(2) x^(2) + ... + a_(96) x^(96)` , then value of `a_(0) - a_(2) + a_(4) - a_(6) + … + a_(96)` is

Options

A. `-1`
B. 0
C. 1
D. 48

Answer

To solve the problem, we need to find the value of the expression \( a_0 - a_2 + a_4 - a_6 + \ldots + a_{96} \) where \( (1 + x + x^2)^{48} = a_0 + a_1 x + a_2 x^2 + \ldots + a_{96} x^{96} \).

### Step 1: Understanding the Expression
We start with the expression \( (1 + x + x^2)^{48} \). The coefficients \( a_n \) represent the coefficients of \( x^n \) in the expansion of this polynomial.

### Step 2: Evaluating at \( x = i \)
To find the sum of the coefficients of even powers with alternating signs, we can evaluate the polynomial at \( x = i \) (where \( i \) is the imaginary unit):
\[ (1 + i + i^2)^{48} \]
Since \( i^2 = -1 \), we have:
\[ 1 + i - 1 = i \]
Thus, we can rewrite it as:
\[ (i)^{48} = i^{48} \]
Since \( i^4 = 1 \), we find:
\[ i^{48} = (i^4)^{12} = 1^{12} = 1 \]

### Step 3: Evaluating at \( x = -i \)
Next, we evaluate the polynomial at \( x = -i \):
\[ (1 - i + (-i)^2)^{48} = (1 - i - 1)^{48} = (-i)^{48} \]
Again using \( (-i)^4 = 1 \):
\[ (-i)^{48} = ((-i)^4)^{12} = 1^{12} = 1 \]

### Step 4: Setting Up the Equations
Now we have two equations:
1. From \( x = i \):
\[ a_0 + a_1 i - a_2 - a_3 i + a_4 + a_5 i - a_6 - a_7 i + \ldots + a_{96} = 1 \]
2. From \( x = -i \):
\[ a_0 - a_1 i - a_2 + a_3 i + a_4 - a_5 i - a_6 + a_7 i + \ldots + a_{96} = 1 \]

### Step 5: Adding the Two Equations
Adding both equations cancels out the terms with \( i \):
\[ 2a_0 - 2a_2 + 2a_4 - 2a_6 + \ldots + 2a_{96} = 2 \]
Dividing by 2 gives:
\[ a_0 - a_2 + a_4 - a_6 + \ldots + a_{96} = 1 \]

### Conclusion
Thus, the value of \( a_0 - a_2 + a_4 - a_6 + \ldots + a_{96} \) is \( \boxed{1} \).