If `(1+x+x)^(2n)=a_(0)+a_(1)x+a_(2)x^(2)+a_(2n)x^(2n)`, then `a_(1)+a_(3)+a_(5)+……..+a_(2n-1)` is equal to

Options

A. `(3^(n)+1)/2`
B. `(3^(n)-1)/2`
C. `(1-3^(n))/2`
D. `(3^(n))/2-1`

Answer

To solve the problem, we need to find the sum of the coefficients of the odd powers of \( x \) in the expansion of \( (1 + x + x^2)^{2n} \).

### Step 1: Define the expression
We start with the expression:
\[ (1 + x + x^2)^{2n} \]
This can be expanded into a polynomial of the form:
\[ a_0 + a_1 x + a_2 x^2 + \ldots + a_{2n} x^{2n} \]

### Step 2: Evaluate at \( x = 1 \)
To find the sum of all coefficients, we set \( x = 1 \):
\[ (1 + 1 + 1^2)^{2n} = (3)^{2n} = 3^{2n} \]
This gives us the total sum of the coefficients:
\[ a_0 + a_1 + a_2 + \ldots + a_{2n} = 3^{2n} \]

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

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

### Step 5: Solve the equations
We can add these two equations to eliminate the odd indexed coefficients:
\[ (a_0 + a_1 + a_2 + \ldots + a_{2n}) + (a_0 - a_1 + a_2 - a_3 + \ldots + a_{2n}) = 3^{2n} + 1 \]
This simplifies to:
\[ 2(a_0 + a_2 + a_4 + \ldots + a_{2n}) = 3^{2n} + 1 \]
From this, we can find the sum of the coefficients of the even powers.

Now, subtract Equation 2 from Equation 1:
\[ (a_0 + a_1 + a_2 + \ldots + a_{2n}) - (a_0 - a_1 + a_2 - a_3 + \ldots + a_{2n}) = 3^{2n} - 1 \]
This simplifies to:
\[ 2(a_1 + a_3 + a_5 + \ldots + a_{2n-1}) = 3^{2n} - 1 \]

### Step 6: Solve for the odd indexed coefficients
Dividing by 2 gives us:
\[ a_1 + a_3 + a_5 + \ldots + a_{2n-1} = \frac{3^{2n} - 1}{2} \]

### Final Answer
Thus, the sum of the coefficients of the odd powers of \( x \) is:
\[ \boxed{\frac{3^{2n} - 1}{2}} \]