anonymous 5 years ago et (1-1/2^2)(1-1/3^2)...(1-1/2010^2)(1-1/2011^2) = x/2*2011. what is the value of x?

$\left(\left(1-\frac{1}{2^2}\right) \left(1-\frac{1}{3^2}\right)\text{...}\right) \left(1-\frac{1}{2010^2}\right) \left(1-\frac{1}{2011^2}\right)==\frac{x 2011}{2}$ I believe you posted the above, however, I think you meant to post: $\left(1-\frac{1}{2^2}\right) \left(1-\frac{1}{3^2}\right)\text{...} \left(1-\frac{1}{2010^2}\right) \left(1-\frac{1}{2011^2}\right)==\frac{x 2011}{2}$ The above can be rewritten as: $\prod _{n=2}^{2011} \left(1-\frac{1}{n^2}\right)=\frac{x 2011}{2}$ The product is equal to: $\prod _{n=2}^{2011} \left(1-\frac{1}{n^2}\right)=\frac{1006}{2011}$ So the equation becomes: $\frac{1006}{2011}=\frac{x 2011}{2}$ Solving for x, $x=\frac{2012}{4044121}$