find n if (n+2)!=2550*n!
Answers
Answered by
62
(n+2)(n+1)n!=2550*n!
(n+2)(n+1)=2550
(n+2)(n+1)=50*51
soo
n=49
(n+2)(n+1)=2550
(n+2)(n+1)=50*51
soo
n=49
Answered by
46
(n+2)! can be written as(n+2)(n+1)n!=2550*n!
(n+2)(n+1)=2550
n^2+2n+n+2
=n^2+3n-2548=0
so it can be either 49 or -51
as n can't be negative so n
=49
(n+2)(n+1)=2550
n^2+2n+n+2
=n^2+3n-2548=0
so it can be either 49 or -51
as n can't be negative so n
=49
Similar questions