MathDB

9

Part of 1986 AMC 12/AHSME

Problems(1)

Series product

Source: AHSME 1986 problem 9

9/30/2011
The product (1−122)(1−132)…(1−192)(1−1102)\left(1 - \frac{1}{2^{2}}\right)\left(1 - \frac{1}{3^{2}}\right)\ldots\left(1 - \frac{1}{9^{2}}\right)\left(1 - \frac{1}{10^{2}}\right) equals
<spanclass=′latex−bold′>(A)</span> 512<spanclass=′latex−bold′>(B)</span> 12<spanclass=′latex−bold′>(C)</span> 1120<spanclass=′latex−bold′>(D)</span> 23<spanclass=′latex−bold′>(E)</span> 710 <span class='latex-bold'>(A)</span>\ \frac{5}{12}\qquad<span class='latex-bold'>(B)</span>\ \frac{1}{2}\qquad<span class='latex-bold'>(C)</span>\ \frac{11}{20}\qquad<span class='latex-bold'>(D)</span>\ \frac{2}{3}\qquad<span class='latex-bold'>(E)</span>\ \frac{7}{10}
AMC