Question -
Answer -
Given:
an = 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 thefirst three terms of the sequence is zero.
Now, let’s check forwhen n = n:
an = 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)}
an =(n – 2)3 – (n – 2)
Here, n – 2 willalways be positive for n > 3
∴ an isalways positive for n > 3