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Telescoping product

Source: 1959 AHSME Problem 37

August 14, 2013
AMCtelescopeProductalgebraAMC 12

Problem Statement

When simplified the product (1−13)(1−14)(1−15)⋯(1−1n)\left(1-\frac13\right)\left(1-\frac14\right)\left(1-\frac15\right)\cdots\left(1-\frac1n\right) becomes: <spanclass=′latex−bold′>(A)</span> 1n<spanclass=′latex−bold′>(B)</span> 2n<spanclass=′latex−bold′>(C)</span> 2(n−1)n<spanclass=′latex−bold′>(D)</span> 2n(n+1)<spanclass=′latex−bold′>(E)</span> 3n(n+1) <span class='latex-bold'>(A)</span>\ \frac1n \qquad<span class='latex-bold'>(B)</span>\ \frac2n\qquad<span class='latex-bold'>(C)</span>\ \frac{2(n-1)}{n}\qquad<span class='latex-bold'>(D)</span>\ \frac{2}{n(n+1)}\qquad<span class='latex-bold'>(E)</span>\ \frac{3}{n(n+1)}
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