# Prove that $\sum_{k=0}^n a_k x^k = 0$ has at least $1$ real root if $\sum_{k=0}^n \frac{a_k}{k+1} = 0$

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.

• Can you think of any way to manipulate $p$ to make that second expression appear? That should be useful, wouldn't you think? – Arthur Jan 9 '18 at 15:54
• ah thank you, i should have copped that thank you – user438263 Jan 9 '18 at 15:55
• 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. – Arthur Jan 9 '18 at 15:57
• Related question : math.stackexchange.com/questions/554262/… – Arnaud D. Sep 4 '18 at 12:03

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.
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 $g(x)$.