Question

para n=2003, o valor de
n! (n²-1)
----------------
(n+1)! é?

Answer 1

Resposta:

n! (n²-1)  

----------------

(n+1)n!

(n²-1)  

----------

(n+1)

=(n+1)(n-1)/(n+1) =n-1 =2003-1=2002

Answer 2

$\frac{n!(n^2-1)}{(n+1)!}~=~\frac{n!.(n+1).(n-1)}{(n+1).(n+1~-1)!}~=~\frac{n!.(n+1).(n-1)}{(n+1).n!}~=~\frac{1~.~1~.~(n-1)}{1~.~1}~=~n-1\\\\\\Substituindo~o~valor~de~''n'':\\\\\\2003-1~=~\boxed{2002}$