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Knowing that $$ \frac{a_0}{1} + \frac{a_1}{2} + \frac{a_2}{3} +\cdots + \frac{a_n}{n+1} =0$$ Prove that $$ a_0 + a_1x + a_2x^2 + \cdots + a_nx^n = 0$$ has at least one real solution.

I suspect that it's proven with the intermediate value theorem. But can't find two numbers that satisfy it.

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  • $\begingroup$ Can you think of any way to manipulate $p$ to make that second expression appear? That should be useful, wouldn't you think? $\endgroup$ – Arthur Jan 9 '18 at 15:54
  • $\begingroup$ ah thank you, i should have copped that thank you $\endgroup$ – user438263 Jan 9 '18 at 15:55
  • $\begingroup$ Well, the hint I was trying to give got spoiled by the answer below. So much for trying to help you find the solution on your own. $\endgroup$ – Arthur Jan 9 '18 at 15:57
  • $\begingroup$ Related question : math.stackexchange.com/questions/554262/… $\endgroup$ – Arnaud D. Sep 4 '18 at 12:03
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Hint : $$\int_0^1 a_0+a_1x+a_2x^2+\ldots a_nx^n= \frac{a_0}{1} + \frac{a_1}{2} + \frac{a_2}{3} +\cdots + \frac{a_n}{n+1}=\color{red}0$$

This integral vanishes, it means that the function was sometimes above the $x$-axis, and sometimes below. Since this function is continuous, it must cut $x$-axis at least once, I.e. must have at least one real solution.

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Let $$g(x)=a_0x+a_1\frac{x^2}2+a_2\frac{x^3}3+\cdots+a_n \frac{x^{n+1}}{n+1}.$$ Apply Rolle's theorem to $g(x)$.

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