Question

A sequence is defined by a_n = n³ − 6n² + 11n − 6, n ϵ 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 ϵ N

$$\text{ For }n=1,a_1=1^3-6\times 1^2+11\times 1-6=0$$

$$\text{ For }n=2,a_2=2^3-6\times 2^2+11\times 2-6=0$$

$$\text{ For }n=3,a_3=3^3-6\times 3^2+11\times 3-6=0$$

$$\text{ For }n=4,a_4=4^3-6\times 4^2+11\times 4-6=6>0$$

$$\text{ For }n=5,a_5=5^3-6\times 5^2+11\times 5-6=24>0$$

$$\text{ and so on }$$

$$\text{ Thus, the first three terms are zero and the rest of the terms are positive in the sequence }.$$