a. n > 1, maka
n! – ( n-1 )! = ( n-1)! (n-1)
b. n > 3, maka
n! – (n-3)! = (n-3)! (n pangkat 3 – 3n pangkat 2 = 2n-1 )
Buktikan bahwa
Jawaban Terkonfirmasi
N! – (n – 1)!
= n(n – 1)! – (n – 1)!
dengan memisalkan (n – 1)! = a
na – a = a(n – 1)
= (n – 1)! (n – 1)
terbukti
==========
n! – (n – 3)!
= n(n – 1)(n – 2)(n – 3)! – (n – 3)!
dengan memisalkan (n – 3)! = a
n(n – 1)(n – 2)a – a = a(n(n – 1)(n – 2) – 1)
= (n – 3)! (n(n – 1)(n – 2) – 1)
= (n – 3)! ((n² – n)(n – 2) – 1)
= (n – 3)! (n³ – 2n² – n² + 2n – 1)
= (n – 3)! (n³ – 3n² + 2n – 1)
terbukti