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Sum of Reciprocals of Triangular Numbers

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April 6, 2013

Problem Statement

Let tn=n(n+1)2t_n=\dfrac{n(n+1)}2 be the nnth triangular number. Find 1t1+1t2+1t3+⋯+1t2002.\dfrac1{t_1}+\dfrac1{t_2}+\dfrac1{t_3}+\cdots+\dfrac1{t_{2002}}. <spanclass=′latex−bold′>(A)</span>40032003<spanclass=′latex−bold′>(B)</span>20011001<spanclass=′latex−bold′>(C)</span>40042003<spanclass=′latex−bold′>(D)</span>40012001<spanclass=′latex−bold′>(E)</span>2<span class='latex-bold'>(A) </span>\dfrac{4003}{2003}\qquad<span class='latex-bold'>(B) </span>\dfrac{2001}{1001}\qquad<span class='latex-bold'>(C) </span>\dfrac{4004}{2003}\qquad<span class='latex-bold'>(D) </span>\dfrac{4001}{2001}\qquad<span class='latex-bold'>(E) </span>2
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