Fórum de Matemática
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prove that if \(a_0,a_1,...,a_{2n}\in R\) and \(\sum_{k=0}^{2n}\frac{a_k}{2n-k+1}=0\) then equation \(a_0x^{2n}+a_1x^{2n-1}+...+a_{2n-1}x+a_0=0\) has at least one solution in \((-1,1)\)

Consider the polynomial \(p(x)=a_0x^{2n}+a_1x^{2n-1}+\cdots +a_{2n-1}x+a_{2n}\). You want to show that \(p(x)=0\) has at least one solution in \((-1,1)\). Take the primitive of \(p\) given by the indefinite integral: \(P(x)=\int_0^x a_0t^{2n}+a_1t^{2n-1}+\cdots +a_{2n-1}t+a_{2n} dt\). It is easy to see that \(P(0)=0\) and \(P(1)=\sum_{k=0}^{2n}\frac{a_k}{2n-k+1}\) which is also zero by hypothesis. Therefore, by Rolle's theorem, there is at least one zero of the derivative of \(P\) (which is \(p\)) in the interval \((0,1)\).

thanks Ruicarpenter