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 \(V_n\) = \(1\over n(n+1)\)

\(\therefore\) \(1\over T_n\) = \(V_n\) – \(V_{n+1}\)

Putting n = 1, 2, 3, …… n

\(\implies\) \(1\over T_1\) + \(1\over T_2\) + \(1\over T_3\) + ….. + \(1\over T_n\) = \(V_1\) – \(V_{n+1}\)

= \({\sum}_{r=1}^{n‎} \)\(1\over T_r\) = \(n^2 + 3n\over 2(n+1)(n+2)\)