A sequence is defined by

Question:

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.

Solution:

Given:

an = 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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