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,

a_n = n³ – 6n² + 11n – 6, n ∈ N 

We can find first three terms of sequence by putting the values of n form 1 to 3. 

When n = 1 : 

a₁ = (1)³ – 6(1)² + 11(1) – 6 

⇒ a₁ = 1 – 6 + 11 – 6 

⇒ a₁ = 12 – 12 

⇒ a₁ = 0 

When 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 

When 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 

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

When n = n : 

a_n = n³ – 6n² + 11n – 6 

⇒ a_n = n³ – 6n² + 11n – 6 – n + n – 2 + 2 

⇒ a_n = n³ – 6n² + 12n – 8 – n + 2 

⇒ a_n = (n)³ – 3×2n(n – 2) – (2)³ – n + 2 

{(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 

∴ a_n is always positive for n > 3.