Answer:
Using n = 1, 2, 3, the first three terms can be calculated.
If n = 1,
a1 = (1)3 – 6(1)2 + 11(1) – 6
a1 = 1 – 6 + 11 – 6
a1 = 12 – 12
a1 = 0
If n = 2,
a2 = (2)3 – 6(2)2 + 11(2) – 6
a2 = 8 – 6(4) + 22 – 6
a2 = 8 – 24 + 22 – 6
a2 = 30 – 30
a2 = 0
If n = 3,
a3 = (3)3 – 6(3)2 + 11(3) – 6
a3 = 27 – 6(9) + 33 – 6
a3 = 27 – 54 + 33 – 6
a3 = 60 – 60
a3 = 0
First three terms of the sequence is zero.
If n = n,
an = n3 – 6n2 + 11n – 6
an = n3 – 6n2 + 11n – 6 – n + n – 2 + 2
an = n3 – 6n2 + 12n – 8 – n + 2
an = (n)3 – 3×2n(n – 2) – (2)3 – n + 2
Using the formula, {(a – b)3 = (a)3 – (b)3 – 3ab(a – b)}
an = (n – 2)3 – (n – 2)
n – 2 will always be positive for n > 3
∴ an is always positive for n > 3