# Formula for even and odd number multiplication

by gaobo9109
Tags: formula, multiplication, number
 P: 463 To find out the product of even numbers think of the factorial. $$n!=1.2.3.4...(n-1)n$$ So you want to find the product $$f(n)=2.4.6.8...(2(n-1))(2n)$$ Notice that this is just the normal factorial function but each number has been multiplied by 2. So it is $$2^n$$ bigger. So it is $$f(n)=2^nn! = 2.4.6.8...(2(n-1))(2n)$$ Next you want to find the product such that $$f(n)=1.3.5.7...(2(n-1)-1)(2n-1)$$ Its similar... divide each one by 2 and $$2^{-n}f(n)=\frac{1}{2}\frac{3}{2}\frac{5}{2}...\frac{2n-3}{2}\frac{n-1}{2}$$ $$2^{-n}f(n)=(1-\frac{1}{2})(3-\frac{1}{2})(5-\frac{1}{2})...((n-1)-\frac{1}{2})(n-\frac{1}{2})$$ $$(n-\frac{1}{2})!=(n-\frac{1}{2})(n-1-\frac{1}{2})...\frac{5}{2}\frac{3}{2}\frac{1}{2}\sqrt{\pi}$$ To work out why this is so look at the definition for the Gamma function. $$\frac{2^n}{\sqrt{\pi}}(n-\frac{1}{2})! = 1.3.5.7...(2n-1)$$ So now you have the formula for both things