Menu Close

solve-1-n-2-n-n-2-4n-

Tinku Tara 0 Comments
Pin




Question Number 171742 by Mikenice last updated on 20/Jun/22
solve: (1+n)!+2(n)!=(n+2)!−4n!
solve:(1+n)!+2(n)!=(n+2)!−4n!
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
(n+2)!−(n+1)!−6n!=0n![(n+1)(n+2)−(n+1)−6]=0⇒n2+3n+2−n−1−6=0⇒n2+2n−5=0⇒n∉Z
Commented by Mikenice last updated on 20/Jun/22
thanks sir
thankssir
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]
(n+1)n!+2n!=(n+2)(n+1)n!−4n!(n+1)n!+2n!−(n+2)(n+1)n!+4n!=0n![(n+1)+2−(n+2)(n+1)+4]=0n![(n+1)(1−n−2)+6]=0n![(n+1)(−n−1)+6]=0either::n!=0nosolutionor::(n+1)(−n−1)+6=0−n2−2n+5=0n2+2n−5=0[n∈rationalnumper]

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *