The sum of odd coefficients in the expansion of `(1+2x-3x^(2))^(1025)` is an odd integer

Answer

To solve the problem of finding the sum of odd coefficients in the expansion of \((1 + 2x - 3x^2)^{1025}\) and determining if it is an odd integer, we can follow these steps:

### Step 1: Find the sum of all coefficients.
To find the sum of all coefficients in the polynomial expansion, we substitute \(x = 1\):

\[ (1 + 2(1) - 3(1)^2)^{1025} = (1 + 2 - 3)^{1025} = (0)^{1025} = 0 \]

### Step 2: Understand the relationship between even and odd coefficients.
In any polynomial expansion, the sum of the coefficients can be expressed as the sum of even coefficients plus the sum of odd coefficients. Let \(S_e\) be the sum of even coefficients and \(S_o\) be the sum of odd coefficients. We have:

\[ S_e + S_o = 0 \]

### Step 3: Use the property of coefficients.
It is a known property that the sum of even coefficients is equal to the sum of odd coefficients when evaluated at \(x = 1\) and \(x = -1\):

\[ S_e = \frac{S(1) + S(-1)}{2} \]
\[ S_o = \frac{S(1) - S(-1)}{2} \]

### Step 4: Calculate \(S(-1)\).
Now, substituting \(x = -1\):

\[ (1 + 2(-1) - 3(-1)^2)^{1025} = (1 - 2 - 3)^{1025} = (-4)^{1025} = -4^{1025} \]

### Step 5: Substitute values into the equations for \(S_e\) and \(S_o\).
From our calculations:
- \(S(1) = 0\)
- \(S(-1) = -4^{1025}\)

Now substituting into the equations for \(S_e\) and \(S_o\):

\[ S_e = \frac{0 + (-4^{1025})}{2} = -2^{2050} \]
\[ S_o = \frac{0 - (-4^{1025})}{2} = \frac{4^{1025}}{2} = 2^{2050} \]

### Step 6: Determine if \(S_o\) is odd.
Since \(S_o = 2^{2050}\), we can see that \(2^{2050}\) is an even number (as it is a power of 2).

### Conclusion:
Thus, the sum of the odd coefficients in the expansion of \((1 + 2x - 3x^2)^{1025}\) is \(2^{2050}\), which is an even integer. Therefore, the statement that the sum of odd coefficients is an odd integer is false.