For a positive integer $n$ and any real number $c$, define $x_k$ recursively by :
\[ x_0=0,x_1=1 \text{ and for }k\ge 0, \;x_{k+2}=\frac{cx_{k+1}-(n-k)x_k}{k+1} \]
Fix $n$ and then take $c$ to be the largest value for which $x_{n+1}=0$. Find $x_k$ in terms of $n$ and $k,\; 1\le k\le n$.
To solve the problem, we need to find \( x_k \) in terms of \( n \) and \( k \) for \( 1 \le k \le n \) given the recursive definition:
\[ x_0 = 0, \quad x_1 = 1, \]
\[ x_{k+2} = \frac{cx_{k+1} - (n-k)x_k}{k+1} \quad \text{for} \; k \ge 0. \]

We are also given that \( x_{n+1} = 0 \) for the largest value of \( c \).

1. **Initial Conditions and Recurrence Relation:**
   - We start with \( x_0 = 0 \) and \( x_1 = 1 \).
   - The recurrence relation is:
     \[ x_{k+2} = \frac{cx_{k+1} - (n-k)x_k}{k+1}. \]

2. **Polynomial Form of \( x_k \):**
   - We assume \( x_k \) is a polynomial in \( c \) of degree \( k-1 \).
   - This polynomial is either purely even or purely odd depending on the parity of \( k-1 \).

3. **Finding \( c \) such that \( x_{n+1} = 0 \):**
   - The equation \( x_{n+1} = 0 \) is an \( n \)-degree polynomial equation in \( c \).
   - The roots of this polynomial are conjectured to be \( c = n+1-2j \) for \( 1 \le j \le n \).

4. **Largest Value of \( c \):**
   - The largest value of \( c \) is \( c = n-1 \).

5. **Verification of \( x_k = \binom{n-1}{k-1} \):**
   - We need to verify that \( x_k = \binom{n-1}{k-1} \) satisfies the recurrence relation when \( c = n-1 \).

6. **Substitution and Simplification:**
   - Substitute \( x_k = \binom{n-1}{k-1} \) into the recurrence relation:
     \[ x_{k+2} = \frac{(n-1)\binom{n-1}{k} - (n-k)\binom{n-1}{k-1}}{k+1}. \]
   - Simplify the expression using properties of binomial coefficients:
     \[ \binom{n-1}{k} = \frac{(n-1)!}{k!(n-1-k)!}, \]
     \[ \binom{n-1}{k-1} = \frac{(n-1)!}{(k-1)!(n-k)!}. \]
   - The recurrence relation becomes:
     \[ x_{k+2} = \frac{(n-1)\frac{(n-1)!}{k!(n-1-k)!} - (n-k)\frac{(n-1)!}{(k-1)!(n-k)!}}{k+1}. \]
   - Simplify further:
     \[ x_{k+2} = \frac{(n-1)!}{(k+1)! (n-1-k)!} = \binom{n-1}{k+1}. \]

7. **Conclusion:**
   - The recurrence relation is satisfied, and thus \( x_k = \binom{n-1}{k-1} \) for \( 1 \le k \le n \).

The final answer is \( \boxed{ x_k = \binom{n-1}{k-1} } \) for \( 1 \le k \le n \).