1. ## Find the Product

I need to find the product:

(1-1/2)(1-1/3)(1-1/4)(1-1/5)......(1-1/499)(1-1/500) =

NOTICE: The list goes from (1-1/2) all the way to
(1 - 1/500).

How do I solve this question?

How do I find the product?

2. Originally Posted by magentarita
I need to find the product:

(1-1/2)(1-1/3)(1-1/4)(1-1/5)......(1-1/499)(1-1/500) =

NOTICE: The list goes from (1-1/2) all the way to
(1 - 1/500).

How do I solve this question?

How do I find the product?
1-1/2 = 1/2

1-1/3 = 2/3

1-1/4 = 3/4

1-1/5 =4/5

.
.
.

1-(1/499) = 498/499

1-(1/500) = 499/500

Multiply all these terms

$\displaystyle \frac{1 * 2 * 3 * 4* 5.....499}{2*3*4*5...499*500}$

cancel common terms in numerator and denominator.

Ans = 1/500

1-1/2 = 1/2

1-1/3 = 2/3

1-1/4 = 3/4

1-1/5 =4/5

.
.
.

1-(1/499) = 498/499

1-(1/500) = 499/500

Multiply all these terms

$\displaystyle \frac{1 * 2 * 3 * 4* 5.....499}{2*3*4*5...499*500}$

cancel common terms in numerator and denominator.

Ans = 1/500
Indeed you do. For n terms this will be

$\displaystyle \frac{(n-1)!}{n!} = \frac{1}{n}$

4. ## tell me

1-1/2 = 1/2

1-1/3 = 2/3

1-1/4 = 3/4

1-1/5 =4/5

.
.
.

1-(1/499) = 498/499

1-(1/500) = 499/500

Multiply all these terms

$\displaystyle \frac{1 * 2 * 3 * 4* 5.....499}{2*3*4*5...499*500}$

cancel common terms in numerator and denominator.

Ans = 1/500
How do I use the formula (1 - n)!/n! = 1/n to find the same answer?

5. ## tell me...

Originally Posted by e^(i*pi)
Indeed you do. For n terms this will be

$\displaystyle \frac{(n-1)!}{n!} = \frac{1}{n}$

How do I use the formula (1 - n)!/n! = 1/n to find the same answer?

6. Originally Posted by magentarita
How do I use the formula (1 - n)!/n! = 1/n to find the same answer?

$\displaystyle =\frac{(500-1)(500-2).....(2)(1)}{(500)(499)...(2)(1)}$

$\displaystyle =\frac{(500-1)!}{500!}$

$\displaystyle =\frac{(499)!}{500!}$

Remember that (n-1)! is not equal to (1-n)!

$\displaystyle =\frac{(n-1)!}{n!} = \frac{1}{n} {\color{red}\ne} \frac{(1-n)!}{n!}$

$\displaystyle =\frac{1}{500}$

7. ## I see...

$\displaystyle =\frac{(500-1)(500-2).....(2)(1)}{(500)(499)...(2)(1)}$

$\displaystyle =\frac{(500-1)!}{500!}$

$\displaystyle =\frac{(499)!}{500!}$

Remember that (n-1)! is not equal to (1-n)!

$\displaystyle =\frac{(n-1)!}{n!} = \frac{1}{n} {\color{red}\ne} \frac{(1-n)!}{n!}$

$\displaystyle =\frac{1}{500}$
I see what I did wrong.