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.