A sequence is defined by $a_{n} = n^{3} - 6 n^{2} + 11 n - 6, n ϵ N$ . Show that the first three terms of the sequence are zero and all other terms are positive.

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Solution

$a_{n} = n^{3} - 6 n^{2} + 11 n - 6, n ϵ N$

The first three terms are $a_{1}, a_{2}$ and $a_{3}$

$a_{1} = (1)^{3} - 6 (1)^{2} + 11 (1) - 6 = 0$

$a_{2} = (2)^{3} - 6 (2)^{2} + 11 (2) - 6 = 0$

$a_{3} = (3)^{3} - 6 (3)^{2} + 11 (3) - 6 = 0$

$∴$ First 1st 3 terms are zero and

$a_{n} = n^{3} - 6 n^{2} + 11 n - 6$

$= (n - 2)^{3} - (n - 2)$ is positive as $n \geq 4$

$∴ a_{n}$ is always positive.

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