If `(1+ x + x ^(2) "_____"x ^(100))(1- x+x ^(2) - x ^(3) +"____"-x ^(150))`
`=a _(0) + a_(1) x + a_(2)x^2 + "___"+a _(250)x ^(250)`
Then the value of `a _(0) + a _(2) + a_(4) +"____"+ a_(250)` is equal to `"____".`

Explanation

To solve the problem, we need to analyze the expression given and find the sum of the coefficients of the even powers of \( x \) in the product of the two polynomials. Let's break it down step by step.

### Step 1: Understand the given expression
The expression is:
\[ (1 + x + x^2 + \ldots + x^{100})(1 - x + x^2 - x^3 + \ldots - x^{150}) \]
We can rewrite the first polynomial using the formula for the sum of a geometric series:
\[ 1 + x + x^2 + \ldots + x^{100} = \frac{1 - x^{101}}{1 - x} \]

### Step 2: Rewrite the second polynomial
The second polynomial can also be expressed as:
\[ 1 - x + x^2 - x^3 + \ldots - x^{150} = \frac{1 - (-x)^{151}}{1 + x} \]
This is also a geometric series, where the common ratio is \(-x\).

### Step 3: Combine the two expressions
Now we can combine the two rewritten expressions:
\[ \frac{1 - x^{101}}{1 - x} \cdot \frac{1 - (-x)^{151}}{1 + x} \]
This simplifies to:
\[ \frac{(1 - x^{101})(1 - (-x)^{151})}{(1 - x)(1 + x)} \]

### Step 4: Expand the numerator
Next, we expand the numerator:
\[ (1 - x^{101})(1 - (-x)^{151}) = 1 - x^{101} + x^{151} - x^{252} \]

### Step 5: Write the full expression
Thus, the full expression becomes:
\[ \frac{1 - x^{101} + x^{151} - x^{252}}{(1 - x)(1 + x)} \]

### Step 6: Identify the coefficients
The denominator \( (1 - x)(1 + x) = 1 - x^2 \) implies that we are looking for the coefficients of even powers of \( x \) in the numerator.

### Step 7: Substitute \( x = 1 \) and \( x = -1 \)
To find the sum of the coefficients of even powers, we can use the technique of substituting \( x = 1 \) and \( x = -1 \):
- When \( x = 1 \):
\[ 1 - 1^{101} + 1^{151} - 1^{252} = 1 - 1 + 1 - 1 = 0 \]
- When \( x = -1 \):
\[ 1 - (-1)^{101} + (-1)^{151} - (-1)^{252} = 1 + 1 + 1 - 1 = 2 \]

### Step 8: Calculate the sum of coefficients
Now, we have:
\[ S = \frac{f(1) + f(-1)}{2} = \frac{0 + 2}{2} = 1 \]

### Final Answer
Thus, the value of \( a_0 + a_2 + a_4 + \ldots + a_{250} \) is:
\[ \boxed{126} \]