If `(1+x+x^(2))^(25)=a_(0)+a_(1)x+a_(2)x^(2)+……………..a_(50)x^(50)`
then the value of
`a_(0)+a_(2)+a_(4)+…………+a_(50)` is

Options

A. odd and of the form 3n
B. odd and of the form `(3n-1)`
C. odd and of the form `(3n+1)`
D. even

Answer

To solve the problem, we need to find the value of \( a_0 + a_2 + a_4 + \ldots + a_{50} \) from the expansion of \( (1 + x + x^2)^{25} \).

### Step 1: Understand the expression
The expression \( (1 + x + x^2)^{25} \) can be expanded using the binomial theorem, which gives us coefficients \( a_n \) for each term \( x^n \).

### Step 2: Evaluate at \( x = 1 \)
First, let's evaluate the expression at \( x = 1 \):
\[ (1 + 1 + 1^2)^{25} = 3^{25} \]
This gives us:
\[ a_0 + a_1 + a_2 + \ldots + a_{50} = 3^{25} \]

### Step 3: Evaluate at \( x = -1 \)
Next, we evaluate the expression at \( x = -1 \):
\[ (1 - 1 + (-1)^2)^{25} = 1^{25} = 1 \]
This gives us:
\[ a_0 - a_1 + a_2 - a_3 + \ldots + a_{50} = 1 \]

### Step 4: Set up the equations
Now we have two equations:
1. \( a_0 + a_1 + a_2 + \ldots + a_{50} = 3^{25} \) (Equation 1)
2. \( a_0 - a_1 + a_2 - a_3 + \ldots + a_{50} = 1 \) (Equation 2)

### Step 5: Add the equations
Adding Equation 1 and Equation 2:
\[ (a_0 + a_1 + a_2 + \ldots + a_{50}) + (a_0 - a_1 + a_2 - a_3 + \ldots + a_{50}) = 3^{25} + 1 \]
This simplifies to:
\[ 2(a_0 + a_2 + a_4 + \ldots + a_{50}) = 3^{25} + 1 \]

### Step 6: Solve for \( a_0 + a_2 + a_4 + \ldots + a_{50} \)
Now, divide both sides by 2:
\[ a_0 + a_2 + a_4 + \ldots + a_{50} = \frac{3^{25} + 1}{2} \]

### Step 7: Final result
Thus, the value of \( a_0 + a_2 + a_4 + \ldots + a_{50} \) is:
\[ \frac{3^{25} + 1}{2} \]