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Possible Duplicate:
Prove existence of a real root.

If $a_0$+$\frac{a_1}{2}$+$\frac{a_2}{3}+\ldots+\frac{a_n}{n+1}=0$, how to prove ${{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+\cdots +{{a}_{n}}{{x}^{n}}=0$ has at least one real root in $(0,1)$.

I know constructor $f(x)={{a}_{0}}x+\frac{{{a}_{1}}}{2}{{x}^{2}}+\frac{{{a}_{2}}}{3}{{x}^{3}}+\cdots +\frac{{{a}_{n}}}{n+1}{{x}^{n+1}}$, and then use the Mean Value Theorem.

I want to know whether we can use mathematical induction to prove, obviously $n = 1$ proposition holds.

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    Is this really a duplicate? The OP seems to ask about possibility of the proof using induction. (Although I have doubts that there is a simple proof using induction.)2012-10-09

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Because $f'(x)={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+\cdots +{{a}_{n}}{{x}^{n}}$ and $f(0)=f(1)=0$ then exists point $x_0 \in (0,\,1)$ such that $f'(x_0)=0$

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Yes.

You just proved it, by the Mean Value Theorem, no? What is an induction needed for?

Let's denote the LHS of the original equation $g(x)$, then your 'constructor' has $f'(x)=g(x)$. So, there is a $\xi\in (0,1)$: $f'(\xi)=\frac{f(1)-f(0)}{1-0} $ But now $f(0)=0$ and, by the hypothesis, also $f(1)=0$. We're done.