If $\frac{a_0}{n+1}+\frac{a_1}{n}+\frac{a_2}{n-1}+\dots +\frac{a_{n-1}}{2}+a_n=0.$ Then the function $f\left(x\right)=a_0{x}^n+a_1{x}^{n-1}+a_2{x}^{n-2}+\dots +a_n$ has in $(0,1)$

Consider the function 

$\varphi \left(x\right)=a_0\frac{x^{n+1}}{n+1}+a_1\frac{x^n}{n}+a_2\frac{x^{n-1}}{n-1}+\dots +a_{n-1}\frac{x^2}{2}+a_{n}x.$

Clearly $\varphi \left(x\right),$ being a polynomial, is continuous on [0, 1] and differentiable on (0, 1)

Also, $\varphi \left(0\right)=0$

and, $\varphi \left(1\right)=\frac{a_0}{n+1}+\frac{a_1}{n}+\frac{a_2}{n-1}+\dots +a_n=0$          [Given]

$\therefore \varphi \left(0\right)=\varphi \left(1\right)$

Thus $\varphi \left(x\right)$ satisfies conditions of Rolle's theorem on [0, 1]

Consequently, there exists $c\in (0,1)$ such that ${\varphi }^{\mathrm{\prime }}\left(c\right)=0$ i.e.

$c\in (0,1)$ is a zero of ${\varphi }^{\mathrm{\prime }}\left(x\right)=a_0{x}^n+a_1{x}^{n-1}+\dots +a_n=f\left(x\right).$