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Challenge · Challenge problem

The nnth triangular number is Tn=n(n+1)2T_n = \tfrac{n(n+1)}{2}. Compute 1T1+1T2+⋯+1T10\dfrac{1}{T_1} + \dfrac{1}{T_2} + \cdots + \dfrac{1}{T_{10}}. The sum is mn\tfrac{m}{n} in lowest terms; find m+nm + n.

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