Question

Calculati ecuatia........♡♡♡♡♡

Answer 1

3 · (n+1)!/2!(n+1-2)! = n+2· n!/(n-2)!
[3 · (n+1)!]/2!(n-1)! = n+2· n!/(n-2)!
[3 · (n-1)!·n·(n+1)]/2(n-1)! = n+[2· (n-2)! ·(n-1)·n]/(n-2)!
[3 · (n-1)!·n·(n+1)]/2(n-1)! = n+[2· (n-2)! ·(n-1)·n]/(n-2)!
3·n·(n+1)/2= n+2·(n-1)·n
(3n^2+3n)/2=n+2n^2-2n
(3n^2+3n)/2=2n^2-n
3n^2+3n=2(2n^2-n)
3n^2+3n=4n^2-2n
3n^2+3n-4n^2+2n=0
-n^2+5n=0
-n(n-5)=0


-n=0 sau n-5=0
n=0 n=5