question_answer

If \[\sum\limits_{r=1}^{n}{\,{{T}_{r}}=\frac{n}{8}\,\,(n+1)\,\,(n+2)\,(n+3)},\] then find \[\sum\limits_{r=1}^{n}{\frac{1}{{{T}_{r}}}}\].

A)  \[\frac{n\,(n+3)}{2\,(n+1)\,\,(n+2)}\]     

B)  \[\frac{n\,(n+1)\,\,(n+2)}{4\,(n+3)}\]

C)  \[\frac{8}{n\,(n+1)\,\,(n+2)\,\,(n+3)}\]

D)  \[\frac{2\,(n+1)\,\,(n+2)}{n\,(n+3)}\]

E)  None of these