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

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Solution

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

Putting $n = 1, 2, 3$ we get

$a_{1} = (1)^{3} - 6 \times 1^{2} + 11 \times 1 - 6 = 1 - 6 + 11 - 6 = 12 - 12 = 0$

$a_{2} = 2^{3} - 6 \times 2^{2} + 11 \times 2 - 6 = 8 - 24 + 22 - 6 = 30 - 30 = 0$

and, $a_{3} = 3^{3} - 6 \times 3^{2} + 11 \times 3 - 6 = 27 - 54 + 33 - 6 = 60 - 60 = 0$

Thus, we have
$a_{1} = a_{2} = a_{3} = 0$

We observe that $a_{n} = n^{3} - 6 n^{2} + 11 n - 6$ is a cubic polynominal in $n$ and it vanished for $n = 1, 2$ and $3$ .

Therefore, by factor theorem $(n - 1), (n - 2)$ and $(n - 3)$ are factors of $a_{n}$ .

Thus, we have
$a_{n} = (n - 1) (n - 2) (n - 3)$
In this expression, if we substitute any value of $n$ which is greater than $3$ , then each factor on the $R H S$ is positive.

Therefore,
$a_{n} > 0$ for all $n > 3$ .
Hence, first three of them sequence are zero and all other terms are positive.

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

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