Express the product (1−1/2^2)(1−1/3^2)(1−1/4^2)···(1−1/21^2) as a fraction a/b, where a and b are relatively prime positive integers.

22/42 = 11/21

How? I can't see the pattern.

1 - 1/k^2 = (k^2-1)/k^2 = (k-1)(k+1)/k^2

write out terms up to n and you will see how the factors cancel.
The product is (n+1)/(2n)

Thank you

To solve this problem, we need to rearrange the given product in a way that allows us to cancel out common terms.

Let's begin by expressing each term in the product as a fraction:

(1 - 1/2^2) = (1 - 1/4) = (3/4)
(1 - 1/3^2) = (1 - 1/9) = (8/9)
(1 - 1/4^2) = (1 - 1/16) = (15/16)
...

We notice that each term has a numerator that is one less than the denominator if written as a single fraction. Therefore, we can represent the nth term as:

(n^2 - 1)/n^2

Now, let's rewrite the given product using this new representation:

(3/4) * (8/9) * (15/16) * ... * ((21^2 - 1)/21^2)

We observe that many terms have a common numerator with the previous denominator. Using this fact, we can simplify the product:

(3/4) * (8/9) * (15/16) * ... * ((21^2 - 1)/21^2)

= (3*8*15*...*(21^2 - 1))/(4*9*16*...*21^2)

Notice that for each term in the numerator, we are multiplying the numerator of the next term in the denominator. This creates several cancellations:

= (3/4) * (1/9) * (1/16) * ... * (1/(21^2))

Hence, the numerator of the fraction is the numerator of the first term, and the denominator is the denominator of the last term:

Numerator: 3
Denominator: 21^2 = 441

Therefore, the product (1−1/2^2)(1−1/3^2)(1−1/4^2)···(1−1/21^2) can be expressed as the fraction 3/441.