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

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

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

Similar to the last question, we may represent the series as $\displaystyle a_n = 3n+2$.

So take two members of this series, $\displaystyle a_n = 3n+2$ and $\displaystyle a_m = 3m+2$. Their product is $\displaystyle a_n * a_m = (3n+2)(3m+2)$. This is supposed to be another member of the series, so let $\displaystyle 3q+2=(3n+2)(3m+2)=9nm+6(n+m)+4$. Solving for q:

$\displaystyle q=3mn+2(n+m)+2/3$. 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

- Feb 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 $\displaystyle b_i$: $\displaystyle (3b_1 +2)(3b_2 + 2)...(3b_x + 2)$. When you multiply this out you get an expansion where all the coefficients are divisible by 3 except for the constant term, which is $\displaystyle 2^x$. If this product is to be represented as a member of the series, then the product must be equal to $\displaystyle 3q +2$ for some integer q. Since all the other coefficients are divisible by 3, we need to show that $\displaystyle 2^x-2$ 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