What is (1-1/4)(1-1/9)(1-1/16)....(1-1/n^2)? - via Mind your decision 2 Responses to Find the product superman had this to say about that: October 14, 2014 at 12:41 am 1-1/n^2 = (n^2-1)/n^2 = (n-1)(n+1)/n^2 (1-1/(n-1)^2)(1-1/n^2) = (n-2)(n)/(n-1)^2 * (n-1)(n+1)/n^2 = (n-2)(n+1)/((n-1)n) Required answer = 1*(n+1)/(2*n) Reply Dinesh had this to say about that: October 14, 2014 at 2:10 am The general term is \{ (1-\frac{1}{n^2}) \} which is the same as \{ \frac{(n-1)(n+1)}{n^2} \}. Separating into two products, we get which after telescopic cancellation gives \{ \frac{N+1}{2N} \} Reply Add Your Comment, Feedback or Opinion Here (Cancel Reply) Your email is safe here. It will not be published or shared. Required fields are marked *My Feedback is: You may use these HTML tags and attributes: <a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <s> <strike> <strong> Name * Email * Website Subscribe to comments

1-1/n^2 = (n^2-1)/n^2 = (n-1)(n+1)/n^2

(1-1/(n-1)^2)(1-1/n^2) = (n-2)(n)/(n-1)^2 * (n-1)(n+1)/n^2 = (n-2)(n+1)/((n-1)n)

Required answer = 1*(n+1)/(2*n)

The general term is \{ (1-\frac{1}{n^2}) \} which is the same as \{ \frac{(n-1)(n+1)}{n^2} \}. Separating into two products, we get

which after telescopic cancellation gives \{ \frac{N+1}{2N} \}