An equation `a_0+a_1x+a_2x^2+...............+a_99x^99+x^100=0` has roots `.^(99)C_0 ,.^(99)C_1,.^(99)C_2....^(99)C_99`. Find the value of `a_99`.

Explanation

To solve the problem, we need to find the value of \( a_{99} \) in the polynomial equation given the roots. Let's break it down step by step.

### Step 1: Understand the Polynomial
The polynomial is given as:
\[ a_0 + a_1 x + a_2 x^2 + \ldots + a_{99} x^{99} + x^{100} = 0 \]
This can be rewritten in standard form:
\[ x^{100} + a_{99} x^{99} + a_{98} x^{98} + \ldots + a_1 x + a_0 = 0 \]

### Step 2: Identify the Roots
The roots of the polynomial are given as:
\[ \binom{99}{0}, \binom{99}{1}, \binom{99}{2}, \ldots, \binom{99}{99} \]
These are the binomial coefficients corresponding to \( n = 99 \).

### Step 3: Use Vieta's Formulas
According to Vieta's formulas, the sum of the roots of the polynomial can be expressed as:
\[ \text{Sum of roots} = -\frac{a_{99}}{1} \]
This means:
\[ \text{Sum of roots} = -a_{99} \]

### Step 4: Calculate the Sum of the Roots
The sum of the binomial coefficients can be calculated using the identity:
\[ \sum_{r=0}^{n} \binom{n}{r} = 2^n \]
For \( n = 99 \):
\[ \sum_{r=0}^{99} \binom{99}{r} = 2^{99} \]

### Step 5: Set Up the Equation
From Step 3 and Step 4, we have:
\[ -a_{99} = 2^{99} \]
This implies:
\[ a_{99} = -2^{99} \]

### Conclusion
Thus, the value of \( a_{99} \) is:
\[ \boxed{-2^{99}} \]