Find product of arimethic sequence in term of x.

[lai001]

My question means that:there is an arithmetic sequence that start from 2x-3 ,and followed by 2x-5,2x-7 and so on, so the diffrence between two terms is -2, and finally what is the product of the terms when reaching the( x-1)th term, in term of x. Please help me to this using the most elementary way as possible.A classmate of mine asked me this question but I searched on the web about this but found less resources or cannot understand. Thanks.

 

[mmm4444bot]

My question means that:there is an arithmetic sequence that start from 2x-3 ,and followed by 2x-5,2x-7 and so on, so the diffrence between two terms is -2, and finally what is the product of the terms when reaching the x-1 term, in term of x. Please help me to this using the most elementary way as possible.A classmate of mine asked me this question but I searched on the web about this but found less resources or cannot understand. Thanks.

Edited: The Gamma Function is not required. I had misunderstood the exercise.

Thanks to Denis, I now understand that the first x-1 terms of the sequence (for any Integer value of x greater than 1) is simply a list of the first x-1 odd numbers, in reverse order.

One expression for the product of the first n odd numbers is:

\(\displaystyle \dfrac{(2n - 1)!}{2^{n-1} \cdot (n-1)!}\)

In your exercise, n = x - 1, so, make that substitution, and you're done.

(In post #9, Denis shows a simpler form, using a double-factorial.)


And, for future reference, an expression for the product of the first n positive, even numbers is:

\(\displaystyle 2^{n} \cdot n!\)

:cool:

 

[lai001]

Did you intend to write "the n-1 term" (above, in red)? That is, you have the first n terms of the sequence, and you would like their product, excluding the nth term.

If this is correct and (2x-3)/(-2) is not a negative Integer or zero, then I think you could express the product using the Gamma Function.

Just that x in the 2x-3 .... and so on
I believe the entire expression can be expressed in term of x as all the variables are in term of x and integers (the first term, number of terms and difference between the term)
If this can expressed in gamma function, please show my how. Thanks.

 

[lai001]

Did you intend to write "the n-1 term" (above, in red)? That is, you have the first n terms of the sequence, and you would like their product, excluding the nth term.If this is correct and (2x-3)/(-2) is not a negative Integer or zero, then I think you could express the product using the Gamma Function.

it is just the same x as the first term above

 

[mmm4444bot]

… all the variables are in [terms] of x and integers

All the variables? I only see one variable (x).

Anyway, you ought to have stated in your original post that x is always an Integer!!

 

[mmm4444bot]

[x-1 term] is just the same x as the first term above

Okay. So n terms multiplied together means x-1 terms multiplied together.

That is, n = x - 1, correct?

 

[mmm4444bot]

Your sequence represents odd numbers from 1 to 2x-3

How did you determine that domain for x? That is, how did you decide that the first term is 1? :?

 

[mmm4444bot]

By using any [Integer] for x

I'm sure you meant "any Integer > 1" for x.

1 is actually the last term...but that's of no importance...

I see now, thanks. I had confused myself, thinking symbol x was being used both as a variable and parameter.

Also, given x, the number of terms is easily determined;
formula: (2x-3 + 1) / 2
Using x = 5 again:
(2*5-3 + 1) / 2 = 4

Or: Number of Terms = x - 1

Turns out that the answer is simply (2x - 3)!!;
!! being the rarely used "double factorial".

Not sure I've seen !! before.

See my edited post #2, for another form. :cool:

 

[lai001]

Edited: The Gamma Function is not required. I had misunderstood the exercise.Thanks to Denis, I now understand that the first x-1 terms of the sequence (for any Integer value of x greater than 1) is simply a list of the first x-1 odd numbers, in reverse order.One expression for the product of the first n odd numbers is:\(\displaystyle \dfrac{(2n - 1)!}{2^{n-1} \cdot (n-1)!}\)In your exercise, n = x - 1, so, make that substitution, and you're done.(In post #9, Denis shows a simpler form, using a double-factorial.)And, for future reference, an expression for the product of the first n positive, even numbers is:\(\displaystyle 2^{n} \cdot n!\):cool:

Thanks!I think this is a useful formula to calculate product of all odd numbers before a number.

 

[mmm4444bot]

You're welcome. I learned something, too.

!!