(1−1/2^2)(1−1/3^2)(1−1/4^2)...(1−1/99^2)(1−1/100^2).
(1−1/2^2)(1−1/3^2)(1−1/4^2)...(1−1/99^2)(1−1/100^2)
= 3/4 * 8/9 * 15/16 * 24/25 * 35/36 * 48/49 * ... * 9999/10000
if we let P(n) be the product of the first n terms
P(1) = 3/4
P(2) = 3/4*8/9 = 2/3
P(3) = 2/3*15/16 = 5/8
P(4) = 5/8*24/25 = 3/5
P(5) = 3/5 * 35/36 = 7/12
P(6) = 7/12 * 48/49 = 4/7
P(7) = 4/7 * 63/64 = 9/16
P(8) = 9/16 * 80/81 = 5/9
p(9) = 5/9 * 99/100 = 11/20
P(10) = 11/20 * 120/121 = 6/11
P(11) = 6/11 * 143/144 = 13/24
...
can you see a pattern ?
since we want P(10000), look at the even n values of P(n)
e.g
P(2) = 2/3
P(4) = 3/5
P(6) = 4/7
P(n) = (n+2)/(2n+2)
so (1−1/2^2)(1−1/3^2)(1−1/4^2)...(1−1/99^2)(1−1/100^2)
= P(10000) = 10002/20002 = 5001/10001
surely there must be a simpler way.
note that we start at n=2, not n=1
That makes the product (n+1)/2n
we can show that by induction
Assuming P(n) = (n+1)/(2n)
then we need to show that P(n+1) = (n+2)/(2(n+1))
So, adding the n+1st factor, we have
P(n+1) = P(n)(1 - 1/(n+1)^2) = (n+1)/(2n)(1 - 1/(n+1)^2)
= (n+1)/(2n)((n+1)^2 - 1)/(n+1)^2
= (n+1)(n^2+2n)/(2n(n+1)^2)
= (n+1)(n+2)*n / (2n(n+1)(n+1))
= (n+2)/(2(n+1))
To find the value of the expression (1−1/2^2)(1−1/3^2)(1−1/4^2)...(1−1/99^2)(1−1/100^2), we can start by simplifying each term.
The expression can be written as follows:
(1 - 1/2^2) * (1 - 1/3^2) * (1 - 1/4^2) * ... * (1 - 1/99^2) * (1 - 1/100^2)
Now let's simplify each term step by step.
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
Continuing this pattern for the remaining terms, we have:
(3/4) * (8/9) * (15/16) * ... * (9801/10000)
Now we can see that the numerator of each fraction is formed by multiplying consecutive odd numbers starting from 3 up to 99^2 (represented by 9801) and the denominator is formed by multiplying consecutive squares of integers starting from 2^2 up to 100^2.
So, the numerator is 3 * 5 * 7 * 9 * ... * 99 * 99, and the denominator is 2^2 * 3^2 * 4^2 * ... * 100^2.
Simplifying further, we can rewrite the numerator and denominator as:
(2 * 3 * 4 * 5 * ... * 99) * 99 / (2^2 * 3^2 * 4^2 * ... * 99^2 * 100)
Now, notice that the numerator and the denominator have the same factors from 2 to 99, with the exception of 99 in the numerator. So, we can cancel out these common factors, resulting in:
99 / (99^2 * 100)
Finally, simplifying further, we have:
1 / (99 * 100)
Therefore, the value of the expression (1−1/2^2)(1−1/3^2)(1−1/4^2)...(1−1/99^2)(1−1/100^2) is 1 / (99 * 100), which can be further simplified to 1 / 9900.