If \({\sum}_{r=1}^{n} T_r\) = \(n\over 8\) (n + 1)(n + 2)(n + 3), then find \({\sum}_{r=1}^{n} \)\(1\over T_r\)
Solution : \(\because\) \(T_n\) = \(S_n – S_{n-1}\) = \({\sum}_{r=1}^{n} T_r\) – \({\sum}_{r=1}^{n – 1} T_r\) = \(n(n+1)(n+2)(n+3)\over 8\) – \((n-1)(n)(n+1)(n+2)\over 8\) = \(n(n+1)(n+2)\over 8\)[(n+3) – (n-1)] = \(n(n+1)(n+2)\over 8\)(4) \(T_n\) = \(n(n+1)(n+2)\over 2\) \(\implies\) \(1\over T_n\) = \(2\over n(n+1)(n+2)\) = \((n+2)-n\over n(n+1)(n+2)\) = \(1\over n(n+1)\) – \(1\over (n+1)(n+2)\) Let \(V_n\) = \(1\over n(n+1)\) \(\therefore\) […]