Proving Cube Numbers Not in Sequence a_n=3n+2

[Natasha1]

Right here is my sequence 2, 5, 8, 11, 14, ...

I have been asked to prove that the cube of any number in the sequence is in the sequence.

my answer:

General term: a_n=3n+2

We need to cube a_n and see if it matches a number in the series i.e. (a_n)^3 = 3q+2 for some integer q.

(a_n)^3
=27n^3 + 54n^2 + 36n + 8
=3(9n^3 + 18n^2 + 12n + 2) +2
=3k+2

If this is a member of the series, then 3q+2 = 3k+2 for some integer q.

Solving for q:

q = k which is always in the sequence.

So the cube of any number is in this sequence.

But now I'm asked to show which cube numbers (therefore not in the sequence, I think ) are not in the sequence and to prove it?

A little confused how to do this one could anyone help please :-)

 

[Fermat]

Check your expansion of (3n+2)^3

(3n+2)^3 = (3n)^3 + 3.(3n)^2.2 + 3.(3n).2^2 + 2^3
(3n+2)^3 = 27n^3 + 54N^2 + 36n + 8

 

[Natasha1]

Check your expansion of (3n+2)^3

(3n+2)^3 = (3n)^3 + 3.(3n)^2.2 + 3.(3n).2^2 + 2^3
(3n+2)^3 = 27n^3 + 54N^2 + 36n + 8

oops thanks ever so much!

 

[Natasha1]

Can anyone see through this one?

 

[Fermat]

Your sequence is a_n = 3n + 2

ergo 3n and 3n+1 are not in the sequence.

Does that help?

 

[Natasha1]

Your sequence is a_n = 3n + 2

ergo 3n and 3n+1 are not in the sequence.

Does that help?

well spotted! merci!

 

[Fermat]

c'est rien!