Simplifying a Product of Sequences

[Euler2718]

Homework Statement

[/B]
Simplify:

[tex] \frac{5\cdot 8\cdot 11 \cdots (3i+2)}{2\cdot 5 \cdot 8 \cdots (3i-1)} [/tex]

Homework Equations

The Attempt at a Solution

I realize the numerator and denominator terms cancel besides the 2, however I'm struggling to write this in a proper form. Only just started sequences, haven't introduced infinite products or sigmas, or anything along those lines. Some insight would be appreciated.

 

[Student100]

Homework Statement

[/B]
Simplify:

[tex] \frac{5\cdot 8\cdot 11 \cdots (3i+2)}{2\cdot 5 \cdot 8 \cdots (3i-1)} [/tex]

Homework Equations

The Attempt at a Solution

I realize the numerator and denominator terms cancel besides the 2, however I'm struggling to write this in a proper form. Only just started sequences, haven't introduced infinite products or sigmas, or anything along those lines. Some insight would be appreciated.

Just the two?

 

[vela]

What do you mean by "proper form"?

 

[Ray Vickson]

Homework Statement

[/B]
Simplify:

[tex] \frac{5\cdot 8\cdot 11 \cdots (3i+2)}{2\cdot 5 \cdot 8 \cdots (3i-1)} [/tex]

Homework Equations

The Attempt at a Solution

I realize the numerator and denominator terms cancel besides the 2, however I'm struggling to write this in a proper form. Only just started sequences, haven't introduced infinite products or sigmas, or anything along those lines. Some insight would be appreciated.

You say you realize that the numerator and denominator terms cancel, but I don't understand what is preventing you from just going ahead and cancelling them.

 

[Euler2718]

Just the two?

I was thinking,

[tex] \frac{3i+2}{2} [/tex]

Because the previous term of the numerator should cancel with the 3i-1 as the pattern suggests.

 

[Student100]

I was thinking,

[tex] \frac{3i+2}{2} [/tex]

Because the previous term of the numerator should cancel with the 3i-1 as the pattern suggests.

Okay, that's what you meant, not ##\frac{1}{2}##. That's it.

 

[Euler2718]

Okay, that's what you meant, not ##\frac{1}{2}##. That's it.

Thanks. Been a bit sick lately; really appreciate the help on this forum.

 

[Student100]

Thanks. Been a bit sick lately; really appreciate the help on this forum.

If you need to prove it to yourself take the first 5 terms of the sequence, and simplify. Then take i = 5 and put it into the expression you just wrote. It'll be the same. What kind of insights were you looking for?

 

[Euler2718]

If you need to prove it to yourself take the first 5 terms of the sequence, and simplify. Then take i = 5 and put it into the expression you just wrote. It'll be the same. What kind of insights were you looking for?

Maybe insights wasn't the proper word. I was at it a while getting no-where so I was hoping for a kick in the right direction, as was the case.

 

[HallsofIvy]

The crucial "insight" is that 3(i- 1)+ 2= 3i- 3+ 2= 3i- 1 so that, yes, the only difference between the sums in the numerator and the denominator is that the denominator starts with "2" that the numerator does not have and that the numerator ends with 3i+ 2 that the denominator does not have. Everything else cancels.