The sum of the co-efficients of all the even powers of x in the expansion of `(2x^2 - 3x + 1)^11` is -

Options

A. `3.6^(10)`
B. `6^(11)`
C. `2^(10).3^(11)`
D. `2^(11).3^(10)`

Answer

To find the sum of the coefficients of all the even powers of \( x \) in the expansion of \( (2x^2 - 3x + 1)^{11} \), we can follow these steps:

### Step 1: Rewrite the expression
We can rewrite the expression as:
\[ (1 - 3x + 2x^2)^{11} \]

### Step 2: Use the Binomial Theorem
According to the Binomial Theorem, the expansion of \( (a + b + c)^n \) can be expressed as:
\[ \sum_{i+j+k=n} \frac{n!}{i!j!k!} a^i b^j c^k \]
where \( i, j, k \) are non-negative integers representing the powers of \( a, b, c \) respectively.

### Step 3: Find the sum of coefficients for \( x = 1 \) and \( x = -1 \)
To find the sum of the coefficients of all terms, we can substitute \( x = 1 \):
\[ (1 - 3(1) + 2(1^2))^{11} = (1 - 3 + 2)^{11} = 0^{11} = 0 \]
This gives us the sum of all coefficients.

Now, substituting \( x = -1 \):
\[ (1 - 3(-1) + 2(-1)^2)^{11} = (1 + 3 + 2)^{11} = 6^{11} \]
This gives us the sum of the coefficients with alternating signs.

### Step 4: Set up equations
Let \( S \) be the sum of the coefficients of even powers of \( x \) and \( T \) be the sum of the coefficients of odd powers of \( x \). We have:
\[ S + T = 0 \quad \text{(from } x = 1\text{)} \]
\[ S - T = 6^{11} \quad \text{(from } x = -1\text{)} \]

### Step 5: Solve the equations
Adding the two equations:
\[ (S + T) + (S - T) = 0 + 6^{11} \]
\[ 2S = 6^{11} \]
\[ S = \frac{6^{11}}{2} = 3 \cdot 6^{10} \]

### Conclusion
Thus, the sum of the coefficients of all even powers of \( x \) in the expansion of \( (2x^2 - 3x + 1)^{11} \) is:
\[ \boxed{3 \cdot 6^{10}} \]