Question Number 171742 by Mikenice last updated on 20/Jun/22
$${solve}:\:\left(\mathrm{1}+{n}\right)!+\mathrm{2}\left({n}\right)!=\left({n}+\mathrm{2}\right)!−\mathrm{4}{n}! \\ $$
Commented by kaivan.ahmadi last updated on 20/Jun/22
![(n+2)!−(n+1)!−6n!=0 n![(n+1)(n+2)−(n+1)−6]=0 ⇒n^2 +3n+2−n−1−6=0⇒ n^2 +2n−5=0⇒n∉Z](Q171746.png)
$$\left({n}+\mathrm{2}\right)!−\left({n}+\mathrm{1}\right)!−\mathrm{6}{n}!=\mathrm{0} \\ $$$${n}!\left[\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)−\left({n}+\mathrm{1}\right)−\mathrm{6}\right]=\mathrm{0} \\ $$$$\Rightarrow{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}−{n}−\mathrm{1}−\mathrm{6}=\mathrm{0}\Rightarrow \\ $$$${n}^{\mathrm{2}} +\mathrm{2}{n}−\mathrm{5}=\mathrm{0}\Rightarrow{n}\notin\mathbb{Z} \\ $$
Commented by Mikenice last updated on 20/Jun/22
$${thanks}\:{sir} \\ $$
Commented by mokys last updated on 21/Jun/22
![(n+1)n! + 2n! = (n+2)(n+1)n! − 4n! (n+1)n! +2n! − (n+2)(n+1)n! + 4n! = 0 n! [( n + 1) +2 − (n+2)(n+1)+4]=0 n! [ (n+1) (1 − n − 2 )+ 6 ] = 0 n! [ (n+1)(−n−1) + 6 ] = 0 either :: n! = 0 no solution or :: (n+1)(−n−1)+6 = 0 −n^2 −2 n + 5 = 0 n^2 +2n−5 = 0 [n∈ rational numper]](Q171805.png)
$$\left({n}+\mathrm{1}\right){n}!\:+\:\mathrm{2}{n}!\:=\:\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right){n}!\:−\:\mathrm{4}{n}!\: \\ $$$$ \\ $$$$\left({n}+\mathrm{1}\right){n}!\:+\mathrm{2}{n}!\:−\:\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right){n}!\:+\:\mathrm{4}{n}!\:=\:\mathrm{0} \\ $$$$ \\ $$$${n}!\:\left[\left(\:{n}\:+\:\mathrm{1}\right)\:+\mathrm{2}\:−\:\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right)+\mathrm{4}\right]=\mathrm{0} \\ $$$$ \\ $$$${n}!\:\left[\:\left({n}+\mathrm{1}\right)\:\left(\mathrm{1}\:−\:{n}\:−\:\mathrm{2}\:\right)+\:\mathrm{6}\:\right]\:=\:\mathrm{0} \\ $$$$ \\ $$$${n}!\:\left[\:\left({n}+\mathrm{1}\right)\left(−{n}−\mathrm{1}\right)\:+\:\mathrm{6}\:\right]\:=\:\mathrm{0} \\ $$$$ \\ $$$${either}\:::\:{n}!\:=\:\mathrm{0}\:\:{no}\:{solution} \\ $$$$ \\ $$$${or}\:::\:\left({n}+\mathrm{1}\right)\left(−{n}−\mathrm{1}\right)+\mathrm{6}\:=\:\mathrm{0} \\ $$$$ \\ $$$$\:−{n}^{\mathrm{2}} \:−\mathrm{2}\:{n}\:+\:\mathrm{5}\:=\:\mathrm{0} \\ $$$$ \\ $$$$\:{n}^{\mathrm{2}} +\mathrm{2}{n}−\mathrm{5}\:=\:\mathrm{0}\:\:\left[{n}\in\:{rational}\:{numper}\right] \\ $$