Question

EN 1998 - se an= (n+1)! - n! / n²[(n-1)! + n! então a1997 é :
Questão envolvendo operações com fatorial , Gabarito: 1÷1998

Answer 1

$Dada \ a \ express\tilde{a}o : \\ \\ a_n \ = \ \frac{(n+1)!-n!}{n^2[(n-1)!+n!]} \\ \\ Vou \ fator\acute{a}-la \ :$

$a_n \ = \ \frac{(n+1)(n)(n-1)!-n(n-1)!}{n^2[(n-1)!+(n).(n-1)!]} \\ a_n \ = \frac{(n)(n-1)![(n+1)-1]}{n^2(n-1)![1+n]} \\ a_n \ = \frac{n+1-1}{n[1+n]} \\ a_n \ = \frac{n}{n[1+n]} \\ \\ \boxed{a_n \ = \frac{1}{1+n}}$


$Ent\tilde{a}o \ \ a_1_9_9_7 \ , \\ \\ a_1_9_9_7 \ = \ \frac{1}{1+1997} \\ a_1_9_9_7 \ = \ \frac{1}{1998} \ \ \ \ \ \ \ \boxed{letra \ b)}$