# How to show that there exists a root of $f(x)=0$?

Let $f(x)=\sum_{k=0}^n a_k x^k$, where $a_k$'s satisfy $\sum_{k=0}^n \frac{a_k}{k+1}=0$.Show that there exists a root of $f(x)=0$ in the interval $(0,1)$.

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I am stuck at first stage. So, I give no work that I have done. –  A.D Jan 1 '13 at 10:30

Hint: Consider $g(x)=\sum_{k=0}^n\frac{a_k}{k+1}x^{k+1}$. Show that $g(0)=g(1)$. Use mean value theorem to infer that $g'(x)=0$ for some $x\in(0,1)$. Now, what is the relation between $g'$ and $f$?
Is $g'=f$ the relationship? –  A.D Jan 1 '13 at 10:45
@A.D Just differentiate $g$ and you'll see. –  user1551 Jan 1 '13 at 11:07
Note that $$\int_0^1 f(x) dx=\sum_{k=0}^n\frac{a_k}{k+1}=0$$ So none of $f(x)>0$ or $f(x)<0$ in whole of $(0,1)$ is possible, so $f$ must change its sign and hence by Intemediate Value Property it must have a root in $(0,1).$