General relation for a pattern?

[BrettJimison]

Homework Statement

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),...

Homework Equations

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 anyone know the general term? Thanks!

 

[Dick]

Homework Statement

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),...

Homework Equations

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 anyone know the general term? Thanks!

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

 

[HallsofIvy]

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!.

 

[BrettJimison]

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!.

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...

 

[HallsofIvy]

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

 

[BrettJimison]

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

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