Show that if $f(x)= \sum\limits_{i=0}^n a_i x^i$ and $a_0+\frac{a_1}{2}+\ldots+\frac{a_n}{n+1}=0$, then there is an $x \in (0,1)$ with $f(x)=0$ Show that if $f(x)= \sum\limits_{i=0}^n a_i x^i$ and $a_0+\dfrac{a_1}{2}+\ldots+\dfrac{a_n}{n+1}=0$, then there is an $x \in (0,1)$ with $f(x)=0.$

you can use the zero point theorem!
if $a_0>0$ and  $\dfrac{a_1}{2}+\ldots+\dfrac{a_n}{n+1}<0$
then the condition you write will turn out to be a critical condition:
$a_0+\dfrac{a_1}{2}+\ldots+\dfrac{a_n}{n+1}=0$
since $f(0)=a_0>0$ and $f(1)=$$-na_0-(n-1)\dfrac{a_1}{2}-\ldots-\dfrac{a_n}{n+1}<0$$\Longrightarrow$$\frac{n-1}{n}\frac{a_{1}}{2}+\frac{n-2}{n}\frac{a_{2}}{3}+......+\frac{1}{n}\frac{a_{n-1}}{n}\le$$\dfrac{a_1}{2}+\ldots+\dfrac{a_n}{n+1}={-a_0}$
a contradiction holds, which imply that $f(0)\cdot{f(1)}\le0$ $,$ then you can use the zero point theorem !
is it helpful ?
