# Simplifying the Expression

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1. Jan 16, 2016

### Morgan Chafe

1. The problem statement, all variables and given/known data

Simplify:

$$\frac{5\cdot 8\cdot 11 \cdots (3i+2)}{2\cdot 5 \cdot 8 \cdots (3i-1)}$$

2. Relevant equations

3. 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.

2. Jan 16, 2016

### Student100

Just the two?

3. Jan 16, 2016

### vela

Staff Emeritus
What do you mean by "proper form"?

4. Jan 16, 2016

### Ray Vickson

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.

5. Jan 16, 2016

### Morgan Chafe

I was thinking,

$$\frac{3i+2}{2}$$

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

6. Jan 16, 2016

### Student100

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

7. Jan 16, 2016

### Morgan Chafe

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

8. Jan 16, 2016

### Student100

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?

9. Jan 16, 2016

### Morgan Chafe

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.

10. Jan 17, 2016

### HallsofIvy

Staff Emeritus
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.