Let `f(x)=x+(1-x)x^3+(1-x)(1-x^2)x^3+.........+(1-x)(1-x^2)........(1-x^(n-1))x^n;(n >= 4)` then :

Options

A. `f (x) =-prod_(r=1)^(n) (1- n ^(r))`
B. `f (x)=1 -prod_(r =1)^(n)(1- x ^(r))`
C. `f (x) =f (x) (prod _(r =1)^(n)(r x ^(r-1))/((1-x ^(r ))))`
D. `f '(x) =f (x) ((prod _(r =1) ^(n) (r x ^(r-4))/((1- x^(r))))`

Answer

To solve the given problem, we need to analyze the function \( f(x) \) defined as:

\[ f(x) = x + (1-x)x^2 + (1-x)(1-x^2)x^3 + \ldots + (1-x)(1-x^2)\ldots(1-x^{n-1})x^n \]

where \( n \geq 4 \).

### Step 1: Rewrite the function in summation form
The function can be expressed in a more compact form. We can observe that each term in the series has a pattern. The \( k \)-th term can be represented as:

\[ (1-x)(1-x^2)(1-x^3)\ldots(1-x^{k-1}) x^k \]

Thus, we can rewrite \( f(x) \) as:

\[ f(x) = \sum_{k=1}^{n} (1-x)(1-x^2)\ldots(1-x^{k-1}) x^k \]

### Step 2: Factor out common terms
Notice that we can factor out \( (1-x) \) from the first two terms, and then continue factoring out \( (1-x^2) \) and so on. This gives us:

\[ f(x) = x + (1-x)x^2 + (1-x)(1-x^2)x^3 + \ldots + (1-x)(1-x^2)\ldots(1-x^{n-1})x^n \]

### Step 3: Simplify the expression
We can express the remaining terms in terms of products of \( (1-x) \):

\[ 1 - f(x) = (1-x)(x^2 + (1-x)x^3 + \ldots + (1-x)(1-x^2)\ldots(1-x^{n-1})x^n) \]

### Step 4: Recognize the pattern in the series
The series can be recognized as a product of terms:

\[ 1 - f(x) = (1-x)(1-x^2)(1-x^3)\ldots(1-x^n) \]

### Step 5: Final expression for \( f(x) \)
Thus, we can conclude that:

\[ f(x) = 1 - \prod_{r=1}^{n} (1-x^r) \]

### Conclusion
The final expression for \( f(x) \) is:

\[ f(x) = 1 - \prod_{r=1}^{n} (1-x^r) \]