Question

(1−1/2^2)(1−1/3^2)(1−1/4^2)...(1−1/99^2)(1−1/100^2).

Answer 1

(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.

Answer 2

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))