how would you write 1 * 3 * 5 *7.... in terms of n?

Question

I need it in terms of n!. For example 1*2*3*4.... is n!. how would you write 1 * 3 * 5 *7.... in terms of n!. (it's being multiplied)

Answer 1

The answer is actually much more complex than one would think! The correct answer though is ...

((2^n)(n-.5)!)/(2(.5!))

That looks kind of messy, so in English that is "2 to the n times n minus 1 half factorial over 2 times 1 half factorial."

Plugging in 1,3,5,7 for n on a graphing calculator, we get 1,3,15,105, which are the correct first four terms in the sequence.

Answer 2

n + 2

Answer 3

General suggestion for finding a pattern. Make a table:

term, value,
k, value(k)
1, 1
2, 3
3, 5
4, 7
now add another column with your idea of what might be happening. in this case, the difference is always 2 so, every time you increment k you add 2.
k, value(k) = 1+2(k-1) = 2k-1, this is exactly the formula for the odd numbers!

Answer 4

quite simple:
all of the numbers are Odd,which means all of them can be written in the form 2n-1,so :

1 * 3 * 5 *.... * (2n-1)
and yes,we write it like:
∞
Π (2n-1) = 1*3*5*7*.....(2n-1) =
n=1
(2n)!
---------
2^n * n!

actually i dont know how "fool" have got this formula (and he/she is the on who deserves the points! ),and im trying to find a way to prove it,but,wow...it just simply works!

Answer 5

2n-1

2 * 1 -1 = 1
2 * 2 -1 = 3
2 * 3 - 1 = 5
etc.

Answer 6

2n-1
So the first term = 2*1-1 = 1
2nd term = 2*2-1 = 3
and so on.

Answer 7

∞
Π 2n-1 = 1*3*5*7*.....
n=1

Answer 8

the answer is 2n!/(2^n*n!)