Question

Me podrian ayudar por favor urgente, con explicación plisss (7-4n )/(3+2n^2 ) 1- (2! )/(3+2n^2 ) ( -1)^(n+1) ( (1 )/(3n^2-1) ) ( (√n +1)/(n^2-1) ) es urgente por favor

Answer 1

Respuesta:

$\frac{-4n+7}{2n^2+3}-\frac{2\left(-1\right)^{n+1}\left(\sqrt{n}+1\right)}{\left(2n^2+3\right)\left(3n^2-1\right)\left(n^2-1\right)}$

Explicación paso a paso:

$\frac{7-4n}{3+2n^2}\cdot \:1-\frac{2!}{3+2n^2}\left(-1\right)^{n+1}\left(\frac{1}{3n^2-1}\right)\left(\frac{\sqrt{n}+1}{n^2-1}\right)$

$\frac{7-4n}{3+2n^2}\cdot \:1-\frac{2!}{3+2n^2}\left(-1\right)^{n+1}\frac{1}{3n^2-1}\cdot \frac{\sqrt{n}+1}{n^2-1}$

$\frac{7-4n}{3+2n^2}\cdot \:1=\frac{7-4n}{3+2n^2}$

$\frac{2!}{3+2n^2}\left(-1\right)^{n+1}\frac{1}{3n^2-1}\cdot \frac{\sqrt{n}+1}{n^2-1}=\frac{2\left(-1\right)^{n+1}\left(\sqrt{n}+1\right)}{\left(3+2n^2\right)\left(3n^2-1\right)\left(n^2-1\right)}$

$\frac{-4n+7}{2n^2+3}-\frac{2\left(-1\right)^{n+1}\left(\sqrt{n}+1\right)}{\left(2n^2+3\right)\left(3n^2-1\right)\left(n^2-1\right)}$

(Espero que te haya servido de ayuda)