Let `n` be a positive integer with `f(n) = 1! + 2! + 3!+.........+n! and p(x),Q(x)` be polynomial in `x` such that `f(n+2)=P(n)f(n+1)+Q(n)f(n)` for all `n >= ,` Then

Answer

To solve the problem, we need to analyze the given function \( f(n) \) and the relationship defined by the polynomials \( P(n) \) and \( Q(n) \).

### Step-by-Step Solution:

1. **Define the function \( f(n) \)**:
\[ f(n) = 1! + 2! + 3! + \ldots + n! \]

2. **Express \( f(n+1) \) and \( f(n+2) \)**:
\[ f(n+1) = 1! + 2! + 3! + \ldots + (n+1)! = f(n) + (n+1)! \]
\[ f(n+2) = 1! + 2! + 3! + \ldots + (n+2)! = f(n+1) + (n+2)! = f(n) + (n+1)! + (n+2)! \]

3. **Substitute into the given relationship**:
The relationship given is:
\[ f(n+2) = P(n)f(n+1) + Q(n)f(n) \]
Substituting the expressions for \( f(n+1) \) and \( f(n+2) \):
\[ f(n) + (n+1)! + (n+2)! = P(n)(f(n) + (n+1)!) + Q(n)f(n) \]

4. **Expand the right-hand side**:
\[ f(n) + (n+1)! + (n+2)! = P(n)f(n) + P(n)(n+1)! + Q(n)f(n) \]
Combine like terms:
\[ f(n) + (n+1)! + (n+2)! = (P(n) + Q(n))f(n) + P(n)(n+1)! \]

5. **Equate coefficients**:
For the coefficients of \( f(n) \):
\[ 1 = P(n) + Q(n) \]
For the coefficients of \( (n+1)! \):
\[ 1 = P(n) \]
Thus, substituting \( P(n) = 1 \) into the first equation:
\[ 1 = 1 + Q(n) \implies Q(n) = 0 \]

6. **Final result**:
Therefore, we conclude:
\[ P(n) = 1 \quad \text{and} \quad Q(n) = 0 \]

### Summary:
The polynomials are:
\[ P(n) = 1, \quad Q(n) = 0 \]