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

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

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

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

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

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

\[\text { and so on }\]

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