# General relation for a pattern?

1. Apr 26, 2015

### BrettJimison

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

I'm in the middle of solving a d.e using the series method. I have come across a weird pattern in part of my solution that I'm confused about:

6, (6)(10),(6)(10)(14),(6)(10)(14)(18),.....

2. Relevant equations

3. The attempt at a solution
I can see its 2(3), 2(3)*2(5), 2(3)*2(5)*2(7)........ But I'm a little confused on what the nth term is. Any help?

Also I can see if we start at n=2, (4n-2),(4n-2)(4n-6),(4n-2)(4n-6)(4n-10),.... But does any one know the general term? Thanks!

2. Apr 26, 2015

### Dick

You could write a general term with the double factorial notation. http://en.wikipedia.org/wiki/Double_factorial

3. Apr 26, 2015

### HallsofIvy

Staff Emeritus
You don't really need to use a special notation like the "double factorial".

I presume you see that it is [(2)(3)][(2)(5)][(2)(7)]...[(2)(2n+1)]. So, first, we have n "2"s : $2^n$.

Then we have (3)(5)(7)(9)...(2n+1), the product of n consecutive odd numbers. We can write that as
$$\frac{2(3)(4)(5)(6)(7)(8)(9)...(2n)(2n+1)}{2(4)(6)(8)...(2n)}$$
The numerator is (2n+1)!. We can write the denominator as $[2(1)][2(2)][2(3)][2(4)]...[2(n)]= 2^n n!$.

4. Apr 26, 2015

### BrettJimison

Thanks for the response hallsofivy,

I'm a little confused: the pattern is 6 , (6)(10) , (6)(10)(14) , (6)(10)(14)(18) , ....

And yes it is also 2(3) , 2(3)*2(5) , 2(3)*2(5)*2(7) ect..... But writing the General term as (2n+1)!/2^n(n)! Doesn't fit the pattern.......

5. Apr 26, 2015

### HallsofIvy

Staff Emeritus
You are missing the $2^n$ in the numerator. You should have
$$2^n\frac{(2n+1)!}{2^n n!}= \frac{(2n+1)!}{n!}$$

6. Apr 26, 2015

### BrettJimison

Ahhh! It's so easy! Thanks, for some reason I just couldn't get it...thanks a lot! It's been a while since I've thought about infinite series