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.

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Solution

Given:
a_n = n³ − 6n² + 11n − 6, n ϵ N
$For n = 1, a_{1} = 1^{3} - 6 \times 1^{2} + 11 \times 1 - 6 = 0 For n = 2, a_{2} = 2^{3} - 6 \times 2^{2} + 11 \times 2 - 6 = 0 For n = 3, a_{3} = 3^{3} - 6 \times 3^{2} + 11 \times 3 - 6 = 0 For n = 4, a_{4} = 4^{3} - 6 \times 4^{2} + 11 \times 4 - 6 = 6 > 0 For n = 5, a_{5} = 5^{3} - 6 \times 5^{2} + 11 \times 5 - 6 = 24 > 0 and so on Thus, the first three terms are zero and the rest of the terms are positive in the sequence .$

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A sequence is defined by an = n3 − 6n2 + 11n − 6, n ϵ N. Show that the first three terms of the sequence are zero and all other terms are positive.

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.