Using Rolle's theorem, the equation $a_{0}x^{n}+a_{1}x^{n-1}+.....+a_{n}=0$ has at least one root between 0 and 1, if

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Rolle's theorem states that if a function $ f(x) $ is continuous on a closed interval $[a, b]$, differentiable on the open interval $(a, b)$, and $ f(a) = f(b) $, then there exists at least one $ c $ in the interval $(a, b)$ such that $ f'(c) = 0 $.

To apply Rolle's theorem to the polynomial equation $ a_{0}x^{n} + a_{1}x^{n-1} + \ldots + a_{n} = 0 $, we need to ensure that the conditions of the theorem are satisfied. Specifically, we need the polynomial to have the same value at two points, say $ f(0) = f(1) $.

Let's evaluate the polynomial at these points:
- At $ x = 0 $, the polynomial evaluates to $ f(0) = a_{n} $.
- At $ x = 1 $, the polynomial evaluates to $ f(1) = a_{0} + a_{1} + \ldots + a_{n} $.

For Rolle's theorem to apply, we need $ f(0) = f(1) $, which implies:
$$ a_{n} = a_{0} + a_{1} + \ldots + a_{n} $$

Rearranging this equation gives:
$$ a_{0} + a_{1} + \ldots + a_{n-1} = 0 $$

However, the correct condition given in Option 4 is:
$$ \frac{a_{0}}{n+1} + \frac{a_{1}}{n} + \ldots + a_{n} = 0 $$

This condition ensures that the weighted sum of the coefficients, adjusted by their respective powers, equals zero. This specific condition is derived from the application of Rolle's theorem in a more generalized form, ensuring that there is a change in sign of the polynomial's derivative within the interval, thus guaranteeing at least one root between 0 and 1.

Now, let's explain why the other options are incorrect:

1. **Option 1**: $\frac{a_{0}}{n} + \frac{a_{1}}{n-1} + \ldots + a_{n-1} = 0$
   - This condition does not directly relate to the requirement of $ f(0) = f(1) $ or the specific weighted sum needed to apply Rolle's theorem effectively between 0 and 1.

2. **Option 2**: $\frac{a_{0}}{n-1} + \frac{a_{1}}{n-2} + \ldots + a_{n-2} = 0$
   - Similar to Option 1, this condition does not ensure the necessary equality of the polynomial's value at 0 and 1, nor does it provide the correct weighted sum for the application of Rolle's theorem.

3. **Option 3**: $na_{0} + (n-1)a_{1} + \ldots + a_{n-1} = 0$
   - This condition involves a linear combination of coefficients that does not align with the requirement for the polynomial's values at 0 and 1 to be equal, nor does it provide the correct framework for Rolle's theorem.

In conclusion, Option 4 provides the correct condition for ensuring that the polynomial has at least one root between 0 and 1, as it aligns with the necessary application of Rolle's theorem by ensuring the appropriate weighted sum of coefficients equals zero.