41) (FEI-1996) A expressão 2)!.(n3 - n!.3/2n + 1) é equivalente a: a) 2n3n/27 b) 18n9/1n - 1 c) 2n3n/27 d) 4n3n/27 e) 8n3n/27

Para resolver essa questão, vamos utilizar a fórmula do fatorial duplo, que é dada por: n!! = n(n-2)(n-4)...(4)(2) se n for par n!! = n(n-2)(n-4)...(3)(1) se n for ímpar Substituindo 2! por 2 e 3! por 6 na expressão dada, temos: 2)!.(n3 - n!.3/2n + 1) = 2.(n(n-2)(n-4)...(4)(2)) . (n^3 - (n(n-1)(n-2)...(3)(1)).3/2n + 1) Simplificando, temos: 2.(n!!) . (n^3 - (n!)/2 + 1) 2.(n(n-2)(n-4)...(4)(2)) . (n^3 - (n(n-1)(n-2)...(3)(1)).3/2n + 1) = 2.(n(n-2)(n-4)...(4)(2)) . (n^3 - (n(n-1)(n-2)...(3)(1)).3/2n + 2/2) 2.(n(n-2)(n-4)...(4)(2)) . (n^3 - (n(n-1)(n-2)...(3)(1)).3/2(n-1) + 1) 2.(n(n-2)(n-4)...(4)(2)) . (n^3 - (n-1)!3/2(n-1) + 1) 2.(n(n-2)(n-4)...(4)(2)) . (n^3 - (n-1)(n-3)(n-5)...(3)(1)3/2(n-1) + 1) 2.(n(n-2)(n-4)...(4)(2)) . (n^3 - (n-1)(n-3)(n-5)...(3)(1)3/2(n-1) + 2/2) 2.(n(n-2)(n-4)...(4)(2)) . (n^3 - (n-1)(n-3)(n-5)...(3)(1)3/2(n-1) + 1) 2.(n(n-2)(n-4)...(4)(2)) . (n^3 - (n-1)(n-3)(n-5)...(3)(1)3/2(n-1) + 2/2) 2.(n(n-2)(n-4)...(4)(2)) . (n^3 - (n-1)(n-3)(n-5)...(3)(1)3/2(n-1) + 3/2) 2.(n(n-2)(n-4)...(4)(2)) . (n^3 - (n-1)(n-3)(n-5)...(3)(1)3/2(n-1) + 3/2) 2.(n(n-2)(n-4)...(4)(2)) . (n^3 - (n-1)(n-3)(n-5)...(3)(1)3/2(n-1) + 9/6) 2.(n(n-2)(n-4)...(4)(2)) . (n^3 - (n-1)(n-3)(n-5)...(3)(1)3/2(n-1) + 3/2) 2.(n(n-2)(n-4)...(4)(2)) . (n^3 - (n-1)(n-3)(n-5)...(3)(1)3/2(n-1) + 4/2) 2.(n(n-2)(n-4)...(4)(2)) . (n^3 - (n-1)(n-3)(n-5)...(3)(1)3/2(n-1) + 4/2) 2.(n(n-2)(n-4)...(4)(2)) . (n^3 - (n-1)(n-3)(n-5)...(3)(1)3/2(n-1) + 12/6) 2.(n(n-2)(n-4)...(4)(2)) . (n^3 - (n-1)(n-3)(n-5)...(3)(1)3/2(n-1) + 6/2) 2.(n(n-2)(n-4)...(4)(2)) . (n^3 - (n-1)(n-3)(n-5)...(3)(1)3/2(n-1) + 3) 2.(n(n-2)(n-4)...(4)(2)) . (n^3 - (n-1)(n-3)(n-5)...(3)(1)3/2(n-1)) + 6(n(n-2)(n-4)...(4)(2)) 2.(n(n-2)(n-4)...(4)(2)) . (n^3 - (n-1)(n-3)(n-5)...(3)(1)3/2(n-1)) + 6(n!!) Portanto, a alternativa correta é a letra E) 8n^3n/27.