Mathematics - Sequence & Series Question with Solution | TestHub

${\mathrm{T}}_r=\frac{1}{1+2+3+\dots +r}=\frac{2}{r(r+1)}$ ${\mathrm{S}}_{10}=2{\sum }_{r=1}^{10}\frac{1}{r(r+1)}=2{\sum }_{r=1}^{10}\left[\frac{r+1}{r(r+1)}-\frac{r}{r(r+1)}\right]$ $=2{\sum }_{r=1}^{10}\left(\frac{1}{r}-\frac{1}{r+1}\right)$ $=2\left[\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\dots +\left(\frac{1}{10}-\frac{1}{11}\right)\right]$ $=2\left[1-\frac{1}{11}\right]=2\times \frac{10}{11}=\frac{20}{11}$