Answer :

Given,


an = 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:


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


{(a – b)3 = (a)3 – (b)3 – 3ab(a – b)}


an = (n – 2)3 – (n – 2)


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


an is always positive for n > 3


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