Question

A sequence is defined by a_n = n³ – 6n² + 11n – 6, n ∈ N. Show that the first three terms of the sequence are zero and all other terms are positive.

Answer 1

Given as

a_n = n³ – 6n² + 11n – 6, n ∈ N

By using the values n = 1, 2, 3 we can find the first three terms.

When n = 1:

a₁ = (1)³ – 6(1)² + 11(1) – 6

= 1 – 6 + 11 – 6

= 12 – 12

= 0

When n = 2:

a₂ = (2)³ – 6(2)² + 11(2) – 6

= 8 – 6(4) + 22 – 6

= 8 – 24 + 22 – 6

= 30 – 30

= 0

When n = 3:

a₃ = (3)³ – 6(3)² + 11(3) – 6

= 27 – 6(9) + 33 – 6

= 27 – 54 + 33 – 6

= 60 – 60

= 0

This shows that the first three terms of the sequence is zero.

Now, let us check for when n = n:

a_n = n³ – 6n² + 11n – 6

= n³ – 6n² + 11n – 6 – n + n – 2 + 2

= n³ – 6n² + 12n – 8 – n + 2

= (n)³ – 3×2n(n – 2) – (2)³ – n + 2

By using the formula, {(a – b)³ = (a)³ – (b)³ – 3ab(a – b)}

a_n = (n – 2)³ – (n – 2)

Here, n – 2 will always be positive for n > 3

Thus, a_n is always positive for n > 3