If $\sum _{r=1}^n t_r=\frac{1}{12}n(n+1)(n+2),$ the value $\sum _{r=1}^n \frac{1}{t_r}$ is

We have $t_r=\sum _{k=1}^r t_k-\sum _{k=1}^{r-1} t_k$

$\begin{array}{l}=\frac{1}{12}r(r+1)(r+2)-\frac{1}{12}(r-1)\left(r\right)(r+1)\\ =\frac{1}{4}r(r+1)\end{array}$

Now, $\frac{1}{t_r}=\frac{4}{r(r+1)}=4\left(\frac{1}{r}-\frac{1}{r+1}\right)$

$\Rightarrow \sum _{r=1}^n \frac{1}{t_r}=4\sum _{r=1}^n \left(\frac{1}{r}-\frac{1}{r+1}\right)=4\left(1-\frac{1}{n+1}\right)=\frac{4n}{n+1}$