Sequence given: 2, 5, 8, 11, 14...

Investigate which products of the sequence members are in the sequence and prove it?

Results 1 to 3 of 3

- February 12th 2006, 01:58 PM #1

- Joined
- Oct 2005
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- February 12th 2006, 04:50 PM #2Originally Posted by
**Natasha**

Similar to the last question, we may represent the series as .

So take two members of this series, and . Their product is . This is supposed to be another member of the series, so let . Solving for q:

. But q must be an integer, and it isn't for any n,m. Thus NO product of members is a member of the series.

-Dan

- February 12th 2006, 08:51 PM #3
I've just had a thought. The question didn't specify how many factors were in the product. You should be able to use the previous post to show that if you take the product of any three members of the series that you get a member of the series.

The key is the constant term in the product. Say we are taking the product of x members of the series, represented by integers : . When you multiply this out you get an expansion where all the coefficients are divisible by 3 except for the constant term, which is . If this product is to be represented as a member of the series, then the product must be equal to for some integer q. Since all the other coefficients are divisible by 3, we need to show that is divisible by 3. This is true of all odd x greater than or equal to 3.

So the product of any 3 members of the series is a member of the series. Specifically, (2)(11)(14) = 308 = 3*102 + 2, so it is a member of the series. Similarly the product of any 5 members of the series is a member: (2)(14)(17)(29)(53) = 731612 = 3*243870 + 2, so it is a member of the series. etc. for 7 members, 9 members...

-Dan