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Exploring the Product: (1-1/2^2) x (1-1/3^2) x ... x (1-1/n^2)
This intriguing product, (1-1/2^2) x (1-1/3^2) x ... x (1-1/n^2), presents an interesting challenge in mathematical exploration. Let's delve into its properties and uncover its fascinating nature.
Simplifying the Expression
First, we can simplify the individual terms in the product:
- (1-1/2^2) = (1 - 1/4) = 3/4
- (1-1/3^2) = (1 - 1/9) = 8/9
- (1-1/n^2) = (n^2 - 1)/n^2 = (n+1)(n-1)/n^2
Notice a pattern emerging. Each term can be expressed as the product of consecutive integers divided by the square of the larger integer.
The Product's General Form
Based on the pattern, we can express the entire product in a more compact form:
(1-1/2^2) x (1-1/3^2) x ... x (1-1/n^2) = (3/4) x (8/9) x (15/16) x ... x ((n^2 - 1)/n^2)
Finding the Product's Value
To determine the product's value, we can observe a remarkable cancellation effect:
- The numerator of each term is the denominator of the next term.
- This cancellation continues until we reach the last term.
Therefore, the product simplifies significantly:
(3/4) x (8/9) x (15/16) x ... x ((n^2 - 1)/n^2) = (3 x 8 x 15 x ... x (n^2 - 1)) / (4 x 9 x 16 x ... x n^2)
Notice how most of the terms cancel out leaving only:
= (3 x (n-1)) / (2 x n)
Conclusion
The product (1-1/2^2) x (1-1/3^2) x ... x (1-1/n^2) simplifies to (3(n-1))/(2n). This expression reveals a fascinating property of this product – its value is determined by the final term 'n' in the sequence.