Question

LES COMPARTO LA SOLUCIÓN A UN PROBLEMA DE ALGEBRA-DIVISION DE POLINOMIOS ​

Answer 1

$\displaystyle\\P(x)=\dfrac{1}{x^2-1}\sum_{i=1}^{2n-1}(x^i-1)^{i+1}\\ \\\\P(x)=\dfrac{1}{x^2-1}\sum_{i=1}^{2n-1}(x-1)^{i+1}(x^{i-1}+x^{i-2}+\cdots+1)^{i+1}\\\\\\P(x)=\dfrac{1}{x+1}\sum_{i=1}^{2n-1}(x-1)^{i}(x^{i-1}+x^{i-2}+\cdots+1)^{i+1}\\\\\\P(x)=\dfrac{1}{x+1}\sum_{i=1}^{2n-1}(x-1)^{i}\left(\sum_{j=0}^{i-1}x^j\right)^{i+1}\\ \\\\r(-1)=\sum_{i=1}^{2n-1}(-2)^{i}\left(\sum_{j=0}^{i-1}(-1)^j\right)^{i+1}$

$\displaystyle\\r(-1)=\sum_{i=1}^{2n-1}(-2)^{i}\cdot\left[\dfrac{1-(-1)^i}{2}\right]^{i+1}\\ \\\\r(-1)=\sum_{k=1}^{n}(-2)^{2k-1}\cdot\left[\dfrac{1-(-1)^{2k-1}}{2}\right]^{2k}\\ \\\\r(-1)=\sum_{k=1}^{n}(-2)^{2k-1}\\ \\ \\r(-1)=-\dfrac{1}{2}\sum_{k=1}^{n}(-2)^{2k}\\ \\ \\r(-1)=-\dfrac{1}{2}\sum_{k=1}^{n}4^{k}\\ \\ \\r(-1)=-\dfrac{1}{2}\cdot \left(\dfrac{4^{n+1}-1}{4-1}-1\right)\\ \\ \\r(-1)=-\dfrac{1}{2}\cdot \left(\dfrac{4^{n+1}-4}{3}\right)\\ \\ \\\boxed{r(-1)=-\dfrac{2}{3}(4^n-1)}$