ChallengeChallenge ShareChallenge · Challenge problemThe nnnth triangular number is Tn=n(n+1)2T_n = \tfrac{n(n+1)}{2}Tn=2n(n+1). Compute 1T1+1T2+⋯+1T10\dfrac{1}{T_1} + \dfrac{1}{T_2} + \cdots + \dfrac{1}{T_{10}}T11+T21+⋯+T101. The sum is mn\tfrac{m}{n}nm in lowest terms; find m+nm + nm+n.Check Show answer Walk through it step by step Practice more like this